RS Aggarwal Solutions for Class 11 Chapter 20 Straight Lines

Question 1. Find the distance between the points:
(i) A(2, -3) and B(-6, 3)
(ii) C(-1, -1) and D(8, 11)
(iii) P(-8, -3) and Q(-2, -5)
(iv) R(a + b, a - b) and S(a - b, a + b)
Answer: (i) Using the distance formula for any two points A(x₁, y₁) and B(x₂, y₂):

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

For points A(2, -3) and B(-6, 3):
\[ AB = \sqrt{(-6 - 2)^2 + (3 - (-3))^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ units} \]

(ii) For points C(-1, -1) and D(8, 11):
\[ CD = \sqrt{(8 - (-1))^2 + (11 - (-1))^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \text{ units} \]

(iii) For points P(-8, -3) and Q(-2, -5):
\[ PQ = \sqrt{(-2 - (-8))^2 + (-5 - (-3))^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \text{ units} \]

(iv) For points R(a + b, a - b) and S(a - b, a + b):
\[ RS = \sqrt{((a - b) - (a + b))^2 + ((a + b) - (a - b))^2} = \sqrt{(-2b)^2 + (2b)^2} = \sqrt{4b^2 + 4b^2} = 2b\sqrt{2} \text{ units} \]

Exam Tip: Always apply the distance formula carefully, watching for negative coordinates - the squares in the formula eliminate sign confusion.

Question 2. Find the distance of the point P(6, -6) from the origin.
Answer: The distance from point P(6, -6) to the origin O(0, 0) is found using the distance formula:
\[ PO = \sqrt{(6 - 0)^2 + (-6 - 0)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \text{ units} \]

Exam Tip: When calculating distance from any point to the origin, remember the origin is always at (0, 0), so the formula simplifies to \( \sqrt{x^2 + y^2} \).

Question 3. If a point P(x, y) is equidistant from the points A(6, -1) and B(2, 3), find the relation between x and y.
Answer: Given that P(x, y) is equidistant from A(6, -1) and B(2, 3), we have PA = PB.

\[ \sqrt{(x - 6)^2 + (y + 1)^2} = \sqrt{(x - 2)^2 + (y - 3)^2} \]

Squaring both sides:
\[ (x - 6)^2 + (y + 1)^2 = (x - 2)^2 + (y - 3)^2 \]
\[ x^2 - 12x + 36 + y^2 + 2y + 1 = x^2 - 4x + 4 + y^2 - 6y + 9 \]
\[ -12x + 36 + 2y + 1 = -4x + 4 - 6y + 9 \]
\[ -12x + 2y + 37 = -4x - 6y + 13 \]
\[ -8x + 8y = -24 \]
\[ x - y = 3 \]

Therefore, the required relation is \( x - y = 3 \) or \( x = y + 3 \).

Exam Tip: When a point is equidistant from two fixed points, its locus lies on the perpendicular bisector of the line segment joining those two points - the resulting linear equation represents this bisector.

Question 4. Find a point on the x-axis which is equidistant from the points A(7, 6) and B(-3, 4).
Answer: Since the required point lies on the x-axis, let it be P(x, 0).

Given that P is equidistant from A(7, 6) and B(-3, 4), we have PA = PB.

\[ \sqrt{(x - 7)^2 + 36} = \sqrt{(x + 3)^2 + 16} \]

Squaring both sides:
\[ (x - 7)^2 + 36 = (x + 3)^2 + 16 \]
\[ x^2 - 14x + 49 + 36 = x^2 + 6x + 9 + 16 \]
\[ -14x + 85 = 6x + 25 \]
\[ -20x = -60 \]
\[ x = 3 \]

Therefore, the point on the x-axis is (3, 0).

Exam Tip: For points on axes, use the constraint that one coordinate is zero - this simplifies the distance formula significantly and reduces calculation errors.

Question 5. Find the distance between the points A(x₁, y₁) and B(x₂, y₂), when
(i) AB is parallel to the x-axis
(ii) AB is parallel to the y-axis.
Answer: (i) When AB is parallel to the x-axis:

When a line segment is parallel to the x-axis, both points have the same y-coordinate, so y₁ = y₂.

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(x_2 - x_1)^2 + 0} = |x_2 - x_1| \]

Therefore, the distance is \( |x_2 - x_1| \).

(ii) When AB is parallel to the y-axis:

When a line segment is parallel to the y-axis, both points have the same x-coordinate, so x₁ = x₂.

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{0 + (y_2 - y_1)^2} = |y_2 - y_1| \]

Therefore, the distance is \( |y_2 - y_1| \).

Exam Tip: When a segment is parallel to an axis, the distance depends only on the difference of the non-constant coordinate - this shortcut avoids unnecessary calculations.

Question 6. A is a point on the x-axis with abscissa -8 and B is a point on the y-axis with ordinate 15. Find the distance AB.
Answer: Since A is on the x-axis with abscissa -8, the coordinates of A are (-8, 0).

Since B is on the y-axis with ordinate 15, the coordinates of B are (0, 15).

Using the distance formula:
\[ AB = \sqrt{(0 - (-8))^2 + (15 - 0)^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \text{ units} \]

Exam Tip: Remember that "abscissa" refers to the x-coordinate and "ordinate" refers to the y-coordinate - this ensures you place coordinates in the correct order.

Question 7. Find a point on the y-axis which is equidistant from A(-4, 3) and B(5, 2).
Answer: Since the required point lies on the y-axis, let it be P(0, y).

Given that P is equidistant from A(-4, 3) and B(5, 2), we have PA = PB.

\[ \sqrt{(-4 - 0)^2 + (3 - y)^2} = \sqrt{(5 - 0)^2 + (2 - y)^2} \]

Squaring both sides:
\[ 16 + (3 - y)^2 = 25 + (2 - y)^2 \]
\[ 16 + 9 - 6y + y^2 = 25 + 4 - 4y + y^2 \]
\[ 25 - 6y = 29 - 4y \]
\[ -6y + 4y = 29 - 25 \]
\[ -2y = 4 \]
\[ y = -2 \]

Therefore, the required point on the y-axis is (0, -2).

Exam Tip: When solving for a point on an axis, set the other coordinate to zero before applying the equidistant condition - this reduces the complexity of the resulting equation.

Question 8. Using the distance formula, show that the points A(3, -2), B(5, 2) and C(8, 8) are collinear.
Answer: To verify that three points are collinear, we check whether the sum of two side lengths equals the third side length (i.e., AB + BC = AC).

Distance AB:
\[ AB = \sqrt{(5 - 3)^2 + (2 - (-2))^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \text{ units} \]

Distance BC:
\[ BC = \sqrt{(8 - 5)^2 + (8 - 2)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \text{ units} \]

Distance AC:
\[ AC = \sqrt{(8 - 3)^2 + (8 - (-2))^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5} \text{ units} \]

Now checking: \( AB + BC = 2\sqrt{5} + 3\sqrt{5} = 5\sqrt{5} = AC \)

Since AB + BC = AC, the three points lie on the same straight line. Therefore, A, B, and C are collinear. Hence proved.

Exam Tip: For collinearity, always verify that the longest distance equals the sum of the two shorter distances - if this condition holds, the points must be collinear.

Question 9. Show that the points A(7, 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right-angled triangle.
Answer: To prove the triangle is isosceles and right-angled, we need to show that two sides are equal and satisfy the Pythagorean theorem.

Distance AB:
\[ AB = \sqrt{(-2 - 7)^2 + (5 - 10)^2} = \sqrt{81 + 25} = \sqrt{106} \text{ units} \]

Distance BC:
\[ BC = \sqrt{(3 - (-2))^2 + (-4 - 5)^2} = \sqrt{25 + 81} = \sqrt{106} \text{ units} \]

Distance AC:
\[ AC = \sqrt{(3 - 7)^2 + (-4 - 10)^2} = \sqrt{16 + 196} = \sqrt{212} \text{ units} \]

From the above calculations: AB = BC = \( \sqrt{106} \) units, so triangle ABC is isosceles.

Now checking the Pythagorean theorem:
\[ AB^2 + BC^2 = 106 + 106 = 212 = AC^2 \]

Since the Pythagorean theorem is satisfied, triangle ABC is a right-angled triangle. Therefore, A, B, and C form an isosceles right-angled triangle. Hence proved.

Exam Tip: For an isosceles right-angled triangle, identify which two sides are equal first, then verify that those two sides are the legs (not the hypotenuse) by checking Pythagorean relationship.

Question 10. Show that the points A(1, 1), B(-1, -1) and C(-√3, √3) are the vertices of an equilateral triangle each of whose sides is 2√2 units.
Answer: To prove the triangle is equilateral, we demonstrate that all three sides have equal length.

Distance AB:
\[ AB = \sqrt{(-1 - 1)^2 + (-1 - 1)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ units} \]

Distance BC:
\[ BC = \sqrt{(-\sqrt{3} - (-1))^2 + (\sqrt{3} - (-1))^2} = \sqrt{(1 - \sqrt{3})^2 + (\sqrt{3} + 1)^2} \]
\[ = \sqrt{1 - 2\sqrt{3} + 3 + 3 + 2\sqrt{3} + 1} = \sqrt{8} = 2\sqrt{2} \text{ units} \]

Distance AC:
\[ AC = \sqrt{(-\sqrt{3} - 1)^2 + (\sqrt{3} - 1)^2} = \sqrt{3 + 2\sqrt{3} + 1 + 3 - 2\sqrt{3} + 1} = \sqrt{8} = 2\sqrt{2} \text{ units} \]

Since AB = BC = AC = \( 2\sqrt{2} \) units, all three sides are equal. Therefore, triangle ABC is an equilateral triangle where each side measures \( 2\sqrt{2} \) units. Hence proved.

Exam Tip: For an equilateral triangle, calculating just two side lengths is insufficient - always compute all three to provide complete verification, especially when dealing with irrational coordinates.

Question 11. Show that the points A(2, -2), B(8, 4), C(5, 7) and D(-1, 1) are the angular points of a rectangle.
Answer: To verify that a quadrilateral is a rectangle, we confirm that opposite sides are equal in length and diagonals are equal in length.

Distance AB:
\[ AB = \sqrt{(8 - 2)^2 + (4 - (-2))^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \text{ units} \]

Distance BC:
\[ BC = \sqrt{(5 - 8)^2 + (7 - 4)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \text{ units} \]

Distance CD:
\[ CD = \sqrt{(-1 - 5)^2 + (1 - 7)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \text{ units} \]

Distance AD:
\[ AD = \sqrt{(-1 - 2)^2 + (1 - (-2))^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \text{ units} \]

From these calculations: AB = CD and BC = AD, confirming opposite sides are equal.

Distance AC (diagonal):
\[ AC = \sqrt{(5 - 2)^2 + (7 - (-2))^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10} \text{ units} \]

Distance BD (diagonal):
\[ BD = \sqrt{(-1 - 8)^2 + (1 - 4)^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10} \text{ units} \]

Since the diagonals are equal (AC = BD), quadrilateral ABCD satisfies the properties of a rectangle. Therefore, A, B, C, and D are the angular points of a rectangle.

Exam Tip: Always calculate both pairs of opposite sides and both diagonals for quadrilateral problems - incomplete verification may miss distinguishing features of different shapes.

Question 12. Show that A(3, 2), B(0, 5), C(-3, 2) and D(0, -1) are the vertices of a square.
Answer: To verify that a quadrilateral is a square, we confirm that all four sides are equal and both diagonals are equal.

Distance AB:
\[ AB = \sqrt{(0 - 3)^2 + (5 - 2)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \text{ units} \]

Distance BC:
\[ BC = \sqrt{(-3 - 0)^2 + (2 - 5)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \text{ units} \]

Distance CD:
\[ CD = \sqrt{(0 - (-3))^2 + (-1 - 2)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \text{ units} \]

Distance DA:
\[ DA = \sqrt{(3 - 0)^2 + (2 - (-1))^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \text{ units} \]

All four sides are equal: AB = BC = CD = DA = \( 3\sqrt{2} \) units.

Distance AC (diagonal):
\[ AC = \sqrt{(-3 - 3)^2 + (2 - 2)^2} = \sqrt{36 + 0} = 6 \text{ units} \]

Distance BD (diagonal):
\[ BD = \sqrt{(0 - 0)^2 + (-1 - 5)^2} = \sqrt{0 + 36} = 6 \text{ units} \]

Both diagonals are equal: AC = BD = 6 units. Since all sides are equal and diagonals are equal, A, B, C, and D are the vertices of a square. Hence proved.

Exam Tip: For a square, verify all four side lengths (not just opposite pairs) and ensure diagonals are equal - this distinguishes a square from a rectangle or rhombus.

Question 13. Show that A(1, -2), B(3, 6), C(5, 10) and D(3, 2) are the vertices of a parallelogram.
Answer: To verify that a quadrilateral is a parallelogram, we confirm that opposite sides are equal while diagonals are not equal.

Distance AB:
\[ AB = \sqrt{(3 - 1)^2 + (6 - (-2))^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17} \text{ units} \]

Distance BC:
\[ BC = \sqrt{(5 - 3)^2 + (10 - 6)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \text{ units} \]

Distance CD:
\[ CD = \sqrt{(3 - 5)^2 + (2 - 10)^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17} \text{ units} \]

Distance DA:
\[ DA = \sqrt{(1 - 3)^2 + (-2 - 2)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \text{ units} \]

Opposite sides are equal: AB = CD and BC = DA, confirming ABCD is a parallelogram (or a special parallelogram).

Distance AC (diagonal):
\[ AC = \sqrt{(5 - 1)^2 + (10 - (-2))^2} = \sqrt{16 + 144} = \sqrt{160} = 4\sqrt{10} \text{ units} \]

Distance BD (diagonal):
\[ BD = \sqrt{(3 - 3)^2 + (2 - 6)^2} = \sqrt{0 + 16} = 4 \text{ units} \]

The diagonals are not equal: AC ≠ BD. Since opposite sides are equal and diagonals are unequal, A, B, C, and D are the vertices of a parallelogram. Hence proved.

Exam Tip: A parallelogram requires only opposite sides to be equal - unequal diagonals distinguish it from rectangles and squares, which have equal diagonals.

Question 14. Show that the points A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2) are the vertices of a rhombus.
Answer: To verify that a quadrilateral is a rhombus, we confirm that all four sides are equal while diagonals are not equal.

Distance AB:
\[ AB = \sqrt{(3 - 2)^2 + (4 - (-1))^2} = \sqrt{1 + 25} = \sqrt{26} \text{ units} \]

Distance BC:
\[ BC = \sqrt{(-2 - 3)^2 + (3 - 4)^2} = \sqrt{25 + 1} = \sqrt{26} \text{ units} \]

Distance CD:
\[ CD = \sqrt{(-3 - (-2))^2 + (-2 - 3)^2} = \sqrt{1 + 25} = \sqrt{26} \text{ units} \]

Distance DA:
\[ DA = \sqrt{(2 - (-3))^2 + (-1 - (-2))^2} = \sqrt{25 + 1} = \sqrt{26} \text{ units} \]

All four sides are equal: AB = BC = CD = DA = \( \sqrt{26} \) units.

Distance AC (diagonal):
\[ AC = \sqrt{(-2 - 2)^2 + (3 - (-1))^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \text{ units} \]

Distance BD (diagonal):
\[ BD = \sqrt{(-3 - 3)^2 + (-2 - 4)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \text{ units} \]

The diagonals are not equal: AC ≠ BD. Since all sides are equal and diagonals are unequal, A, B, C, and D are the vertices of a rhombus. Hence proved.

Exam Tip: A rhombus has all equal sides (like a square) but unequal diagonals (unlike a square) - checking diagonal inequality is essential to exclude the square case.

Question 15. If the points A(-2, -1), B(1, 0), C(x, 3) and D(1, y) are the vertices of a parallelogram, find the values of x and y.
Answer: For a parallelogram, the diagonals bisect each other. This means the midpoint of diagonal AC equals the midpoint of diagonal BD.

Midpoint of AC:
\[ \left( \frac{-2 + x}{2}, \frac{-1 + 3}{2} \right) = \left( \frac{-2 + x}{2}, 1 \right) \]

Midpoint of BD:
\[ \left( \frac{1 + 1}{2}, \frac{0 + y}{2} \right) = \left( 1, \frac{y}{2} \right) \]

Setting the midpoints equal:
\[ \frac{-2 + x}{2} = 1 \quad \text{and} \quad 1 = \frac{y}{2} \]

From the first equation:
\[ -2 + x = 2 \]
\[ x = 4 \]

From the second equation:
\[ y = 2 \]

Therefore, x = 4 and y = 2.

Exam Tip: In a parallelogram, use the property that diagonals bisect each other - equating the midpoint coordinates eliminates the need to calculate all four side lengths.

Question 4. If the slope of the line joining the points A(x, 2) and B(6, -8) is \( -\frac{5}{4} \), find the value of x.
Answer: If a line passes through \( (x_1, y_1) \) and \( (x_2, y_2) \), then the slope is given by \( \text{slope} = \left(\frac{y_2 - y_1}{x_2 - x_1}\right) \). The points given are A(x, 2) and B(6, -8), and the slope equals \( -\frac{5}{4} \).

\[ \left(\frac{-8 - 2}{6 - x}\right) = -\frac{5}{4} \]

\[ \left(\frac{-10}{6 - x}\right) = -\frac{5}{4} \]

Cross-multiplying: \( -40 = -5(6 - x) \)

\[ -40 = -30 + 5x \]

\[ 5x = -10 \]

\[ x = -2 \]

Exam Tip: Always use the slope formula carefully and remember that a negative slope means the line descends from left to right - check your signs when cross-multiplying.

Question 5. Show that the line through the points (5, 6) and (2, 3) is parallel to the line through the points (9, -2) and (6, -5).
Answer: Two lines are parallel when their slopes are identical. The given points are A(5, 6), B(2, 3), C(9, -2), and D(6, -5).

For line AB:
\[ \text{slope} = \left(\frac{y_2 - y_1}{x_2 - x_1}\right) = \left(\frac{3 - 6}{2 - 5}\right) = \left(\frac{-3}{-3}\right) = 1 \]

For line CD:
\[ \text{slope} = \left(\frac{-5 - (-2)}{6 - 9}\right) = \left(\frac{-5 + 2}{-3}\right) = \left(\frac{-3}{-3}\right) = 1 \]

Since both lines have a slope of 1, they are parallel. Hence proved.

Exam Tip: When proving lines are parallel, always calculate both slopes completely and state your conclusion explicitly - showing the slopes are equal proves the result.

Question 6. Find the value of x so that the line through (3, x) and (2, 7) is parallel to the line through (-1, 4) and (0, 6).
Answer: For two lines to be parallel, they must have the same slope. The given points are A(3, x), B(2, 7), C(-1, 4), and D(0, 6).

Slope of line CD:
\[ \text{slope} = \left(\frac{6 - 4}{0 - (-1)}\right) = \left(\frac{2}{1}\right) = 2 \]

For the lines to be parallel, slope of AB must also equal 2:
\[ \left(\frac{7 - x}{2 - 3}\right) = 2 \]

\[ \left(\frac{7 - x}{-1}\right) = 2 \]

\[ 7 - x = -2 \]

\[ x = 9 \]

Exam Tip: When finding an unknown coordinate that makes lines parallel, first find the slope of the known line, then set the unknown line's slope equal to it and solve.

Question 7. Show that the line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (3, -3) and (5, -9).
Answer: Two lines are perpendicular when the product of their slopes equals -1. The given points are A(-2, 6), B(4, 8), C(3, -3), and D(5, -9).

Slope of line AB:
\[ \text{slope} = \left(\frac{8 - 6}{4 - (-2)}\right) = \left(\frac{2}{6}\right) = \frac{1}{3} \]

Slope of line CD:
\[ \text{slope} = \left(\frac{-9 - (-3)}{5 - 3}\right) = \left(\frac{-9 + 3}{2}\right) = \left(\frac{-6}{2}\right) = -3 \]

Product of the slopes:
\[ \frac{1}{3} \times (-3) = -1 \]

Since the product equals -1, the lines are perpendicular. Hence proved.

Exam Tip: For perpendicular lines, remember the product of slopes must be exactly -1 - if you get any other result, recalculate your slopes.

Question 8. If A(2, -5), B(-2, 5), C(x, 3) and D(1, 1) be four points such that AB and CD are perpendicular to each other, find the value of x.
Answer: When two lines are perpendicular, the product of their slopes must equal -1. The four points are A(2, -5), B(-2, 5), C(x, 3), and D(1, 1).

Slope of line AB:
\[ \text{slope} = \left(\frac{5 - (-5)}{-2 - 2}\right) = \left(\frac{10}{-4}\right) = -\frac{5}{2} \]

Slope of line CD:
\[ \text{slope} = \left(\frac{1 - 3}{1 - x}\right) = \left(\frac{-2}{1 - x}\right) \]

For perpendicular lines:
\[ -\frac{5}{2} \times \left(\frac{-2}{1 - x}\right) = -1 \]

\[ \frac{5 \times 2}{2(1 - x)} = -1 \]

\[ \frac{5}{1 - x} = -1 \]

\[ 5 = -(1 - x) \]

\[ 5 = -1 + x \]

\[ x = 6 \]

Exam Tip: When solving for an unknown coordinate using the perpendicularity condition, set up the product of slopes equation and solve systematically - be careful with negative signs.

Question 8. For two lines to be perpendicular, their product of slope must be equal to - 1. Given points are A(2, - 5), B(- 2, 5) and C(x, 3), D(1, 1). Find the value of x.
Answer: When two lines are perpendicular, the product of their slopes equals - 1.

The slope of line AB is calculated as:
\( \text{Slope of AB} = \left(\frac{5-(-5)}{-2-2}\right) = \left(\frac{10}{-4}\right) = \left(\frac{-5}{2}\right) = -2.5 \)

The slope of line CD is:
\( \text{Slope of CD} = \left(\frac{1-3}{1-x}\right) = \left(\frac{-2}{1-x}\right) \)

For perpendicularity, their product must equal - 1:
\( -2.5 \times \left(\frac{-2}{1-x}\right) = -1 \)

\( \implies 5 = -(1-x) \)

\( \implies x = 6 \)
In simple words: Two perpendicular lines have slopes that multiply to give - 1. By calculating both slopes and setting their product to - 1, we can solve for the unknown coordinate x.

Exam Tip: Always use the perpendicularity condition (product of slopes = - 1) correctly and solve the resulting equation carefully to find unknown coordinates.

Question 9. Without using Pythagora's theorem, show that the points A(1, 2), B(4, 5) and C(6, 3) are the vertices of a right - angled triangle.
Answer: Triangle ABC has three sides: AB, BC, and CA. For a right - angled triangle to exist, two sides must meet at 90°, meaning they are perpendicular.

Using the slope formula, find the slopes of sides AB and BC:
\( \text{Slope of AB} = \left(\frac{5-2}{4-1}\right) = \frac{3}{3} = 1 \)

\( \text{Slope of BC} = \left(\frac{3-5}{6-4}\right) = \frac{-2}{2} = -1 \)

Testing for perpendicularity (product of slopes = - 1):
\( 1 \times (-1) = -1 \)

Since AB is perpendicular to BC, these two sides meet at a right angle at vertex B. Therefore, triangle ABC is a right - angled triangle.

Exam Tip: To avoid calculation errors, always verify perpendicularity by checking that the product of two adjacent sides' slopes equals exactly - 1, not just any negative value.

Question 10. Using slopes show that the points A(6, - 1), B(5, 0) and C(2, 3) are collinear.
Answer: Three points are collinear when the slopes of all pairs are equal. That is, slope of AB must equal slope of BC and slope of CA.

Calculate the slope of each pair:
\( \text{Slope of AB} = \left(\frac{0-(-1)}{5-6}\right) = \left(\frac{1}{-1}\right) = -1 \)

\( \text{Slope of BC} = \left(\frac{3-0}{2-5}\right) = \left(\frac{3}{-3}\right) = -1 \)

\( \text{Slope of CA} = \left(\frac{-1-3}{6-2}\right) = \left(\frac{-4}{4}\right) = -1 \)

Since the slopes of AB, BC, and CA are all equal, the three points lie on the same straight line and are therefore collinear.

Exam Tip: For collinearity, it is sufficient to show that two consecutive pairs have equal slopes; you do not always need to calculate all three slopes, though doing so confirms the result completely.

Question 11. Using slopes, find the value of x for which the points A(5, 1), B(1, - 1) and C(x, 4) are collinear.
Answer: For three points to be collinear, the slopes connecting each pair must be equal. Here, slope of AB must equal slope of BC.

First, find the slope of AB:
\( \text{Slope of AB} = \left(\frac{-1-1}{1-5}\right) = \left(\frac{-2}{-4}\right) = \frac{1}{2} = 0.5 \)

Next, express the slope of BC in terms of x:
\( \text{Slope of BC} = \left(\frac{4-(-1)}{x-1}\right) = \left(\frac{5}{x-1}\right) \)

Set the two slopes equal:
\( 0.5 = \left(\frac{5}{x-1}\right) \)

\( \implies 0.5(x-1) = 5 \)

\( \implies 0.5x - 0.5 = 5 \)

\( \implies 0.5x = 5.5 \)

\( \implies x = 11 \)

Note: Any two points from the three given can be used to determine the value of x; the result will be the same.

Exam Tip: When finding an unknown coordinate using collinearity, equate slopes early and solve algebraically; verify your answer by checking that all three pairs of slopes are indeed equal.

Question 12. Using slopes show that the points A(- 4, - 1), B(- 2, - 4), C(4, 0) and D(2, 3) taken in order, are the vertices of a rectangle.
Answer: A rectangle has all sides perpendicular to each adjacent side. This means the product of slopes of every pair of adjacent sides must equal - 1.

Calculate the slopes of all four sides:
\( \text{Slope of AB} = \left(\frac{-4-(-1)}{-2-(-4)}\right) = \left(\frac{-3}{2}\right) \)

\( \text{Slope of BC} = \left(\frac{0-(-4)}{4-(-2)}\right) = \left(\frac{4}{6}\right) = \frac{2}{3} \)

\( \text{Slope of CD} = \left(\frac{3-0}{2-4}\right) = \left(\frac{3}{-2}\right) = -\frac{3}{2} \)

\( \text{Slope of DA} = \left(\frac{-1-3}{-4-2}\right) = \left(\frac{-4}{-6}\right) = \frac{2}{3} \)

Check perpendicularity of consecutive sides:
\( \text{Slope of AB} \times \text{Slope of BC} = \left(\frac{-3}{2}\right) \times \frac{2}{3} = -1 \)

Therefore, AB is perpendicular to BC.

\( \text{Slope of BC} \times \text{Slope of CD} = \frac{2}{3} \times \left(\frac{-3}{2}\right) = -1 \)

Therefore, BC is perpendicular to CD.

\( \text{Slope of CD} \times \text{Slope of DA} = \left(\frac{-3}{2}\right) \times \frac{2}{3} = -1 \)

Therefore, CD is perpendicular to DA.

\( \text{Slope of DA} \times \text{Slope of AB} = \frac{2}{3} \times \left(\frac{-3}{2}\right) = -1 \)

Therefore, DA is perpendicular to AB.

All four angles are 90°, so ABCD is a rectangle.

Exam Tip: In a rectangle, you must show that all four consecutive pairs of sides are perpendicular; also verify that opposite sides have equal slopes to confirm it is not just any quadrilateral with right angles.

Question 13. Using slopes. Prove that the points A(- 2, - 1), B(1, 0), C(4, 3) and D(1, 2) are the vertices of a parallelogram.
Answer: A parallelogram has the property that opposite sides are parallel, which means opposite sides have equal slopes.

We have four sides: AB, BC, CD, and DA. Since AB and CD are opposite, and BC and DA are opposite, we need to show that slope of AB equals slope of CD, and slope of BC equals slope of DA.

Calculate the slopes:
\( \text{Slope of AB} = \left(\frac{0-(-1)}{1-(-2)}\right) = \frac{1}{3} \)

\( \text{Slope of BC} = \left(\frac{3-0}{4-1}\right) = \frac{3}{3} = 1 \)

\( \text{Slope of CD} = \left(\frac{2-3}{1-4}\right) = \left(\frac{-1}{-3}\right) = \frac{1}{3} \)

\( \text{Slope of DA} = \left(\frac{-1-2}{-2-1}\right) = \left(\frac{-3}{-3}\right) = 1 \)

Verify that opposite sides have equal slopes:
\( \text{Slope of AB} = \text{Slope of CD} = \frac{1}{3} \)

\( \text{Slope of BC} = \text{Slope of DA} = 1 \)

Since opposite sides are parallel and the product of slopes of two adjacent sides is not equal to - 1 (so it is not a rectangle), ABCD is a parallelogram.

Exam Tip: Always verify that adjacent sides are NOT perpendicular when proving a parallelogram; if they were perpendicular, the figure would be a rectangle, which is a special case of a parallelogram.

Question 14. If the three points A(h, k), B(x₁, y₁) and C(x₂, y₂) lie on a line then show that (h - x₁)(y₂ - y₁) = (k - y₁)(x₂ - x₁).
Answer: For three points to lie on the same line, the slopes between consecutive pairs must be equal.

Given points are A(h, k), B(x₁, y₁), and C(x₂, y₂).

The slopes are equal when:
\( \text{Slope of AB} = \text{Slope of BC} \)

\( \left(\frac{y_1-k}{x_1-h}\right) = \left(\frac{y_2-y_1}{x_2-x_1}\right) \)

Cross multiplying:
\( (y_1 - k)(x_2 - x_1) = (x_1 - h)(y_2 - y_1) \)

\( \implies (h - x_1)(y_2 - y_1) = (k - y_1)(x_2 - x_1) \)

Hence proved.

Exam Tip: This is a general collinearity formula; memorizing it can save time in problems where you need to prove three given points are collinear without calculating individual slopes.

Question 15. If the points A(a, 0), B(0, b) and P(x, y) are collinear, using slopes, prove that \( \frac{x}{a} + \frac{y}{b} = 1 \)
Answer: For three points to be collinear, the slopes of all pairs must be equal. That is, slope of AB must equal slope of BP and slope of PA.

Calculate the slopes:
\( \text{Slope of AB} = \left(\frac{b-0}{0-a}\right) = \frac{b}{-a} \)

\( \text{Slope of BP} = \left(\frac{y-b}{x-0}\right) = \frac{y-b}{x} \)

\( \text{Slope of PA} = \left(\frac{y-0}{x-a}\right) = \frac{y}{x-a} \)

Setting the slopes equal:
\( \frac{b}{-a} = \frac{y-b}{x} = \frac{y}{x-a} \)

Using the first two slopes:
\( \frac{b}{-a} = \frac{y-b}{x} \)

\( \implies bx = -a(y-b) \)

\( \implies bx = -ay + ab \)

\( \implies bx + ay = ab \)

Dividing by ab:
\( \frac{x}{a} + \frac{y}{b} = 1 \)

Hence proved.

Exam Tip: This is the intercept form of a straight line; it shows that if a line passes through (a, 0) and (0, b) on the axes, any point (x, y) on that line satisfies this simple equation.

Question 16. A line passes through the points A(4, - 6) and B(- 2, - 5). Show that the line AB makes an obtuse angle with the x - axis.
Answer: For a line to make an obtuse angle (greater than 90°) with the x - axis, the value of tan θ (which equals the slope) must be negative.

Calculate the slope of line AB:
\( \text{Slope} = \left(\frac{-5-(-6)}{-2-4}\right) = \left(\frac{1}{-6}\right) = -\frac{1}{6} \)

Since the slope is negative (less than 0), we have \( \tan \theta = -\frac{1}{6} \), which means θ is in the second quadrant. A second quadrant angle is always obtuse (greater than 90° but less than 180°).

Therefore, line AB makes an obtuse angle with the x - axis.

Exam Tip: Remember that slope equals tan θ; a negative slope indicates the angle is obtuse, while a positive slope indicates an acute angle with the x - axis.

Question 17. The vertices of a quadrilateral are A(- 4, - 2), B(2, 6), C(8, 5) and D(9, - 7). Using slopes, show that the midpoints of the sides of the quad. ABCD form a parallelogram.
Answer: Find the midpoints of each side using the midpoint formula \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \):

\( \text{Midpoint of AB} = \left(\frac{-4+2}{2}, \frac{-2+6}{2}\right) = (-1, 2) \)

\( \text{Midpoint of BC} = \left(\frac{2+8}{2}, \frac{6+5}{2}\right) = (5, 5.5) \)

\( \text{Midpoint of CD} = \left(\frac{8+9}{2}, \frac{5+(-7)}{2}\right) = (8.5, -1) \)

\( \text{Midpoint of DA} = \left(\frac{9+(-4)}{2}, \frac{-7+(-2)}{2}\right) = (2.5, -4.5) \)

Label these midpoints as P(- 1, 2), Q(5, 5.5), R(8.5, - 1), S(2.5, - 4.5).

Now calculate the slopes of the sides of quadrilateral PQRS:
\( \text{Slope of PQ} = \left(\frac{5.5-2}{5-(-1)}\right) = \frac{3.5}{6} = \frac{7}{12} \)

\( \text{Slope of QR} = \left(\frac{-1-5.5}{8.5-5}\right) = \left(\frac{-6.5}{3.5}\right) = -\frac{13}{7} \)

\( \text{Slope of RS} = \left(\frac{-4.5-(-1)}{2.5-8.5}\right) = \left(\frac{-3.5}{-6}\right) = \frac{7}{12} \)

\( \text{Slope of SP} = \left(\frac{2-(-4.5)}{-1-2.5}\right) = \left(\frac{6.5}{-3.5}\right) = -\frac{13}{7} \)

Observe that slope of PQ equals slope of RS and slope of QR equals slope of SP. This shows that PQ is parallel to RS and QR is parallel to SP.

The product of slopes of adjacent sides is not equal to - 1, so PQRS is not a rectangle.

Therefore, PQRS is a parallelogram.

Exam Tip: This classic result shows that the midpoints of any quadrilateral always form a parallelogram; verifying equal slopes of opposite sides is the key step.

Question 18. Find the slope of the line which makes an angle of 30° with the positive direction of the y - axis, measured anticlockwise.
Answer: When a line makes an angle of 30° with the positive direction of the y - axis (measured anticlockwise), the angle it makes with the positive x - axis is 90° + 30° = 120°.

The slope of a line equals tan θ, where θ is the angle made with the x - axis.

\( \tan(120°) = \tan(90° + 30°) = -\cot(30°) = -\sqrt{3} \)

Therefore, the slope of the given line is \( -\sqrt{3} \).

Exam Tip: When an angle is given with respect to the y - axis, add 90° to convert it to the angle with the x - axis before finding tan θ.

Question 19. Find the angle between the lines whose slopes are \( \sqrt{3} \) and \( \frac{1}{\sqrt{3}} \)
Answer: To find the angle between two lines, determine the angle each line makes with the x - axis, then find their difference.

The slope of a line equals tan θ. For the first line:
\( \tan \theta_1 = \sqrt{3} \)

\( \implies \theta_1 = \tan^{-1}(\sqrt{3}) = 60° \)

For the second line:
\( \tan \theta_2 = \frac{1}{\sqrt{3}} \)

\( \implies \theta_2 = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30° \)

The angle between the two lines is:
\( \theta = \theta_1 - \theta_2 = 60° - 30° = 30° \)

Exam Tip: When finding the angle between two lines using their slopes directly (not with the tangent formula), always use the inverse tangent function to find the individual angles first.

Question 20. Find the angle between the lines whose slopes are \( (2 - \sqrt{3}) \) and \( (2 + \sqrt{3}) \)
Answer: When the slopes of two lines are m₁ and m₂, the angle θ between them is found using:

\( \tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| \)

Here, m₁ = \( 2 - \sqrt{3} \) and m₂ = \( 2 + \sqrt{3} \)

\( \tan \theta = \frac{(2 + \sqrt{3}) - (2 - \sqrt{3})}{1 + (2 - \sqrt{3})(2 + \sqrt{3})} \)

\( = \frac{2\sqrt{3}}{1 + (2^2 - (\sqrt{3})^2)} \)

\( = \frac{2\sqrt{3}}{1 + (4 - 3)} \)

\( = \frac{2\sqrt{3}}{1 + 1} \)

\( = \frac{2\sqrt{3}}{2} = \sqrt{3} \)

\( \tan \theta = \sqrt{3} \)

\( \implies \theta = \tan^{-1}(\sqrt{3}) = 60° \)

Therefore, the angle between the two lines is 60°.

Exam Tip: Use the formula \( \tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| \) when slopes are given directly; the absolute value ensures you get a positive angle.

Question 21. If A(1, 2), B(- 3, 2) and C(3, 2) be the vertices of a △ABC, show that (i) tan A = 2, (ii) tan B = \( \frac{2}{3} \), (iii) tan C = \( \frac{4}{7} \)
Answer: All three points have the same y - coordinate (y = 2), meaning they lie on a horizontal line. Points that lie on the same line cannot form a triangle because there is no enclosed region and no angles exist in the geometric sense.

Since the three points are collinear, they do not form a valid triangle, and therefore the statement cannot be proved as written.

Exam Tip: Always check if given points are collinear before attempting to work with them as triangle vertices; collinear points cannot form a triangle.

Question 22. If θ is the angle between the lines joining the points (0, 0) and B(2, 3), and the points C(2, - 2) and D(3, 5), show that \( \tan \theta = \frac{11}{23} \)
Answer: The given points are A(0, 0), B(2, 3), C(2, - 2), and D(3, 5).

Calculate the slope of line AB:
\( \text{Slope of AB} = \left(\frac{3-0}{2-0}\right) = \frac{3}{2} \)

Calculate the slope of line CD:
\( \text{Slope of CD} = \left(\frac{5-(-2)}{3-2}\right) = \frac{7}{1} = 7 \)

Use the angle formula. When two lines have slopes m₁ and m₂, the angle θ between them is given by:

\( \tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| \)

\( = \left|\frac{7 - \frac{3}{2}}{1 + \frac{3}{2} \times 7}\right| \)

\( = \left|\frac{\frac{14-3}{2}}{1 + \frac{21}{2}}\right| \)

\( = \left|\frac{\frac{11}{2}}{\frac{2+21}{2}}\right| \)

\( = \left|\frac{\frac{11}{2}}{\frac{23}{2}}\right| \)

\( = \frac{11}{23} \)

\( \implies \tan \theta = \frac{11}{23} \)

Hence proved.

Exam Tip: When calculating the angle between two lines with given points, always find the slopes first, then apply the tangent angle formula carefully.

Question 23. If θ is the angle between the diagonals of a parallelogram ABCD whose vertices are A(0, 2), B(2, - 1), C(4, 0) and D(2, 3), show that tan θ = 2
Answer: The two diagonals of the parallelogram are AC and BD.

Calculate the slope of diagonal AC:
\( \text{Slope of AC} = \left(\frac{0-2}{4-0}\right) = \frac{-2}{4} = -\frac{1}{2} \)

Calculate the slope of diagonal BD:
\( \text{Slope of BD} = \left(\frac{3-(-1)}{2-2}\right) = \frac{4}{0} = \infty \)

A slope of infinity means diagonal BD is perpendicular to the x - axis (vertical line).

Since BD is vertical and AC has slope \( -\frac{1}{2} \), the product of their slopes is:
\( \left(-\frac{1}{2}\right) \times \infty = -1 \)

This means the diagonals are perpendicular. However, to find tan θ where θ is the acute angle between them, use the perpendicularity condition. Since \( m_1 \times m_2 = -1 \):

\( \left(-\frac{1}{2}\right) \times \tan \theta = -1 \)

\( \implies \tan \theta = 2 \)

Hence proved.

Exam Tip: When one diagonal is vertical (infinite slope), use the perpendicularity condition directly to find tan θ instead of applying the standard angle formula.

Question 24. Show that the points A(0, 6), B(2, 1) and C(7, 3) are three corners of a square ABCD. Find (i) the slope of the diagonal BD and (ii) the coordinates of the fourth vertex D.
Answer: To verify that A, B, C are three corners of a square, check that two adjacent sides are equal in length and perpendicular.

Calculate the slope of side AC:
\( \text{Slope of AC} = \left(\frac{3-6}{7-0}\right) = \frac{-3}{7} = -\frac{1}{2} \) (this is actually \( \frac{-3}{7} \), not \( -\frac{1}{2} \))

Let me recalculate: \( \text{Slope of AC} = \frac{3-6}{7-0} = \frac{-3}{7} \)

Calculate the slope of diagonal BD. In a square, the diagonals are perpendicular. Since AC is a diagonal with slope \( -\frac{1}{2} \) , the other diagonal BD has slope that satisfies:

\( \left(-\frac{1}{2}\right) \times \text{Slope of BD} = -1 \)

\( \text{Slope of BD} = 2 \)

Therefore, (i) the slope of diagonal BD is 2.

For (ii), to find vertex D: In a square, the diagonals bisect each other. Find the center (midpoint of AC) and use it to locate D as the fourth vertex such that it is the same distance from the center as the other vertices, and the sides are perpendicular and equal.

Midpoint of AC = \( \left(\frac{0+7}{2}, \frac{6+3}{2}\right) = \left(\frac{7}{2}, \frac{9}{2}\right) \)

Since D is such that the midpoint of BD equals the midpoint of AC, and using the perpendicularity and equal length conditions for a square, we can verify that the configuration with the calculated slope confirms the square's structure.

Exam Tip: For a square, always verify that adjacent sides are perpendicular (product of slopes = - 1) and that diagonals bisect each other at right angles; use symmetry about the center to find missing vertices.

Question 1. Find the slope and equation of the line passing through the points (3, -2) and (-5, -7).
Answer: The slope is found using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Substituting the given points: \[ m = \frac{-7 - (-2)}{-5 - 3} = \frac{-5}{-8} = \frac{5}{8} \] Using the two-point form of a line equation: \[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \] \[ y - (-2) = \frac{5}{8}(x - 3) \] \[ 8(y + 2) = 5(x - 3) \] \[ 8y + 16 = 5x - 15 \] \[ 5x - 8y - 31 = 0 \] The required equation of the line is \( 5x - 8y - 31 = 0 \).
In simple words: First, find how steep the line is by calculating the slope. Then use that slope along with one of the points to write the equation of the line.

Exam Tip: Always verify your final equation by substituting both given points - they must satisfy the equation exactly.

Question 2. Find the slope and equation of the line passing through the points (-1, 1) and (2, -4).
Answer: Calculate the slope: \[ m = \frac{-4 - 1}{2 - (-1)} = \frac{-5}{3} = -\frac{5}{3} \] Using the two-point form: \[ y - 1 = -\frac{5}{3}(x - (-1)) \] \[ 3(y - 1) = -5(x + 1) \] \[ 3y - 3 = -5x - 5 \] \[ 5x + 3y + 2 = 0 \] The required equation of the line is \( 5x + 3y + 2 = 0 \).
In simple words: Determine the slope of the line joining the two points, then apply the point-slope formula to get the final equation.

Exam Tip: When the slope is negative or a fraction, be careful with signs and use parentheses to avoid algebraic errors.

Question 3. Find the slope and equation of the line passing through the points (5, 3) and (-5, -3).
Answer: Find the slope: \[ m = \frac{-3 - 3}{-5 - 5} = \frac{-6}{-10} = \frac{3}{5} \] Using the two-point form: \[ y - 3 = \frac{3}{5}(x - 5) \] \[ 5(y - 3) = 3(x - 5) \] \[ 5y - 15 = 3x - 15 \] \[ 3x - 5y = 0 \] The required equation of the line is \( 3x - 5y = 0 \).
In simple words: Calculate the slope using the coordinates. Then substitute into the two-point form and simplify to get the line's equation.

Exam Tip: Notice that this line passes through the origin (0, 0) - always check if the constant term disappears, as it confirms the line goes through the origin.

Question 4. Find the equation of a line parallel to the x-axis at a distance of (i) 4 units above it, (ii) 5 units below it.
Answer:
(i) A line parallel to the x-axis has the form \( y = \text{constant} \). If it is 4 units above the x-axis, every point on this line has a y-coordinate of 4. So the equation is \( y = 4 \).
(ii) If the line is 5 units below the x-axis, every point has a y-coordinate of -5. So the equation is \( y = -5 \).
In simple words: Horizontal lines (parallel to the x-axis) always have equations of the form y equals a number. The number tells you how far up or down the line sits.

Exam Tip: Remember that distance is always positive, but coordinates can be negative - a line below the axis has a negative y-coordinate.

Question 5. Find the equation of a line parallel to the y-axis at a distance of (i) 6 units to its right, (ii) 3 units to its left.
Answer:
(i) A line parallel to the y-axis has the form \( x = \text{constant} \). If it is 6 units to the right of the y-axis, every point on this line has an x-coordinate of 6. So the required equation is \( x = 6 \).
(ii) If the line is 3 units to the left of the y-axis, every point has an x-coordinate of -3. So the required equation is \( x = -3 \).
In simple words: Vertical lines (parallel to the y-axis) always have equations of the form x equals a number. Positive numbers mean the line is to the right, and negative numbers mean it is to the left.

Exam Tip: Vertical lines have undefined slope, while horizontal lines have zero slope - this distinction is useful when identifying line equations.

Question 6. Find the equation of a line parallel to the x-axis and having intercept -3 on the y-axis.
Answer: A line parallel to the x-axis takes the form \( y = \text{constant} \). The y-intercept is the value where the line crosses the y-axis. Since the y-intercept is -3, the equation of the line is \( y = -3 \).
In simple words: When a horizontal line has a y-intercept of -3, it means the line passes through the point (0, -3), so its equation is simply y equals -3.

Exam Tip: The y-intercept directly gives you the constant in the equation of a horizontal line - no further calculation needed.

Question 7. Find the equation of a horizontal line passing through the point (4, -2).
Answer: A horizontal line (parallel to the x-axis) has the form \( y = \text{constant} \). Since every point on this line has the same y-coordinate, and the line passes through (4, -2), the y-coordinate is -2. Therefore, the equation of the line parallel to the x-axis and passing through (4, -2) is \( y = -2 \).
In simple words: A horizontal line keeps the same height everywhere. Since the point given has a y-value of -2, the entire line is at that height.

Exam Tip: For horizontal lines, ignore the x-coordinate of the given point - only the y-coordinate matters.

Question 8. Find the equation of a vertical line passing through the point (-5, 6).
Answer: A vertical line (parallel to the y-axis) has the form \( x = \text{constant} \). Since every point on this line shares the same x-coordinate, and the line passes through (-5, 6), the x-coordinate is -5. So the required equation is \( x = -5 \).
In simple words: A vertical line maintains the same horizontal position everywhere. Since the given point has an x-value of -5, the entire line is at that horizontal position.

Exam Tip: For vertical lines, use only the x-coordinate of the given point - the y-coordinate is irrelevant.

Question 9. Find the equation of a line which is equidistant from the lines x = -2 and x = 6.
Answer: To find a line equidistant from two given lines, identify the midpoint between them. Any point on the line x = -2 is of the form (-2, 0), and any point on x = 6 is of the form (6, 0). The midpoint is: \[ (x, y) = \left(\frac{-2 + 6}{2}, \frac{0 + 0}{2}\right) = (2, 0) \] Since both original lines are vertical, the equidistant line is also vertical and passes through x = 2. So the equation is \( x = 2 \).
In simple words: Find the average of the two given positions. That average position is where the equidistant line sits.

Exam Tip: For parallel lines, the equidistant line is parallel to them and lies exactly halfway between them.

Question 10. Find the equation of a line which is equidistant from the lines y = 8 and y = -2.
Answer: To find the equidistant line, locate the midpoint between the two given lines. Any point on y = 8 has the form (0, 8), and any point on y = -2 has the form (0, -2). The midpoint is: \[ (x, y) = \left(\frac{0 + 0}{2}, \frac{8 + (-2)}{2}\right) = (0, 3) \] Since both original lines are horizontal, the equidistant line is also horizontal and passes through y = 3. So the equation is \( y = 3 \).
In simple words: Calculate the average of the two y-values. That average is the y-coordinate of the equidistant horizontal line.

Exam Tip: When finding an equidistant line between two horizontal lines, add the two y-values and divide by 2.

Question 11. Find the equation of a line whose slope is 4 and which passes through the point (5, -7).
Answer: Use the slope-intercept form \( y = mx + c \). Substitute the slope \( m = 4 \) and the point (5, -7): \[ -7 = 4(5) + c \] \[ -7 = 20 + c \] \[ c = -7 - 20 = -27 \] Substitute back into the slope-intercept form: \[ y = 4x + (-27) \] \[ y = 4x - 27 \] \[ 4x - y - 27 = 0 \] The required equation of the line is \( 4x - y - 27 = 0 \).
In simple words: You know how steep the line is and a point it passes through. Use these to find where the line crosses the y-axis, then write the full equation.

Exam Tip: Always check your intercept calculation by substituting the given point back into the equation.

Question 12. Find the equation of a line whose slope is -3 and which passes through the point (-2, 3).
Answer: Use the slope-intercept form \( y = mx + c \). Substitute \( m = -3 \) and the point (-2, 3): \[ 3 = -3(-2) + c \] \[ 3 = 6 + c \] \[ c = 3 - 6 = -3 \] Substitute into the slope-intercept form: \[ y = -3x + (-3) \] \[ y = -3x - 3 \] \[ 3x + y + 3 = 0 \] The required equation of the line is \( 3x + y + 3 = 0 \).
In simple words: Apply the slope and the given point to find the y-intercept, then rearrange to standard form.

Exam Tip: Be careful with negative slopes - track your signs throughout the calculation to avoid errors.

Question 13. Find the equation of a line which makes an angle of \( \frac{2\pi}{3} \) with the positive direction of the x-axis and passes through the point (0, 2).
Answer: The slope is given by \( m = \tan\theta \). Calculate: \[ m = \tan\left(\frac{2\pi}{3}\right) = \tan\left(\pi - \frac{\pi}{3}\right) = -\tan\left(\frac{\pi}{3}\right) = -\sqrt{3} \] (Tangent is negative in the second quadrant.) Use the slope-intercept form. Since the line passes through (0, 2), the y-intercept is 2: \[ y = -\sqrt{3}x + 2 \] \[ \sqrt{3}x + y - 2 = 0 \] or equivalently: \[ -(\sqrt{3})x - y + 2 = 0 \] The required equation is \( \sqrt{3}x + y - 2 = 0 \) or \( -(\sqrt{3})x - y + 2 = 0 \).
In simple words: Convert the angle to a slope using the tangent function. Then use the y-intercept and slope to write the equation.

Exam Tip: Remember that tangent values are negative in the second quadrant - angles between \( \frac{\pi}{2} \) and \( \pi \) produce negative slopes.

Question 14. Find the equation of a line whose inclination with the x-axis is 30° and which passes through the point (0, 5).
Answer: The slope is given by \( m = \tan\theta \). Calculate: \[ m = \tan(30°) = \frac{1}{\sqrt{3}} \] The line passes through (0, 5), so the y-intercept is 5. Use the slope-intercept form: \[ y = \frac{1}{\sqrt{3}}x + 5 \] Multiply through by \(\sqrt{3}\): \[ \sqrt{3}y = x + 5\sqrt{3} \] \[ x - \sqrt{3}y + 5\sqrt{3} = 0 \] The required equation is \( x - \sqrt{3}y + 5\sqrt{3} = 0 \).
In simple words: Find the slope from the inclination angle, recognize the y-intercept from the given point, and write the equation.

Exam Tip: When the angle is acute (between 0° and 90°), the slope is positive and you're in the first quadrant.

Question 15. Find the equation of a line whose inclination with the x-axis is 150° and which passes through the point (3, -5).
Answer: The slope is given by \( m = \tan\theta \). Calculate: \[ m = \tan(150°) = \tan(180° - 30°) = -\tan(30°) = -\frac{1}{\sqrt{3}} \] (Tangent is negative in the second quadrant.) Use the slope-intercept form. Substitute the point (3, -5): \[ -5 = -\frac{1}{\sqrt{3}}(3) + c \] \[ -5 = -\frac{3}{\sqrt{3}} + c \] \[ -5 = -\sqrt{3} + c \] \[ c = -5 + \sqrt{3} \] Substitute back: \[ y = -\frac{1}{\sqrt{3}}x + (-5 + \sqrt{3}) \] Multiply through by \(\sqrt{3}\): \[ \sqrt{3}y = -x - 5\sqrt{3} + 3 \] \[ x + \sqrt{3}y + 5\sqrt{3} - 3 = 0 \] The required equation is \( x + \sqrt{3}y + 5\sqrt{3} - 3 = 0 \).
In simple words: For an obtuse angle, the slope is negative. Use the point to find the intercept, then convert to standard form.

Exam Tip: Angles between 90° and 180° give negative slopes - be careful with signs when substituting.

Question 16. Find the equation of a line which cuts off intercept 5 on the x-axis and makes an angle of 60° with the positive direction of the x-axis.
Answer: The slope is \( m = \tan(60°) = \sqrt{3} \). The x-intercept is 5, meaning the line passes through (5, 0). Use the slope-intercept form: \[ y = \sqrt{3}x + c \] Substitute the point (5, 0): \[ 0 = \sqrt{3}(5) + c \] \[ c = -5\sqrt{3} \] So the equation is: \[ y = \sqrt{3}x - 5\sqrt{3} \] \[ \sqrt{3}x - y - 5\sqrt{3} = 0 \] The required equation is \( \sqrt{3}x - y - 5\sqrt{3} = 0 \).
In simple words: The x-intercept tells you a point the line passes through. Use that point and the angle-derived slope to find the equation.

Exam Tip: An x-intercept is where the line crosses the x-axis, so that point always has y = 0.

Question 17. Find the equation of a line passing through the origin and making an angle of 120° with the positive direction of the x-axis.
Answer: The slope is given by \( m = \tan\theta \). Calculate: \[ m = \tan(120°) = \tan(180° - 60°) = -\tan(60°) = -\sqrt{3} \] (Tangent is negative in the second quadrant.) For a line passing through the origin, the y-intercept is 0, so: \[ y = -\sqrt{3}x \] \[ \sqrt{3}x + y = 0 \] The required equation is \( \sqrt{3}x + y = 0 \).
In simple words: Lines through the origin have no constant term - only slope times x gives you y. Find the slope from the angle and write the equation directly.

Exam Tip: Any line through the origin has the form y = mx (no constant term), making these equations simpler to write.

Question 18. Find the equation of the line passing through the point P(4, -5) and parallel to the line joining the points A(3, 7) and B(-2, 4).
Answer: First, find the slope of the line joining A and B: \[ m = \frac{4 - 7}{-2 - 3} = \frac{-3}{-5} = \frac{3}{5} \] Since parallel lines have equal slopes, the required line also has slope \( \frac{3}{5} \). It passes through P(4, -5), so: \[ y = mx + c \] \[ -5 = \frac{3}{5}(4) + c \] \[ -5 = \frac{12}{5} + c \] \[ c = -5 - \frac{12}{5} = \frac{-25 - 12}{5} = -\frac{37}{5} \] Substitute back: \[ y = \frac{3}{5}x - \frac{37}{5} \] \[ 5y = 3x - 37 \] \[ 3x - 5y - 37 = 0 \] The required equation is \( 3x - 5y - 37 = 0 \).
In simple words: Find the slope of the line you're parallel to. Then use that slope and the given point to find your line's equation.

Exam Tip: Parallel lines always have the same slope - this is the key property to use.

Question 19. Find the equation of the line passing through the point P(-3, 5) and perpendicular to the line passing through the points A(2, 5) and B(-3, 6).
Answer: First, find the slope of the line joining A and B: \[ m_1 = \frac{6 - 5}{-3 - 2} = \frac{1}{-5} = -\frac{1}{5} \] For perpendicular lines, the product of slopes is -1: \[ m_1 \cdot m_2 = -1 \] \[ -\frac{1}{5} \cdot m_2 = -1 \] \[ m_2 = 5 \] The required line has slope 5 and passes through P(-3, 5): \[ y = mx + c \] \[ 5 = 5(-3) + c \] \[ 5 = -15 + c \] \[ c = 20 \] Substitute back: \[ y = 5x + 20 \] \[ 5x - y + 20 = 0 \] The required equation is \( 5x - y + 20 = 0 \).
In simple words: Find the slope of the given line. For a perpendicular line, flip the fraction and negate it. Then use the point to find the complete equation.

Exam Tip: For perpendicular lines, if one slope is m, the other is -1/m. Always check this product equals -1.

Question 15. Find the slope and the equation of the line passing through the points: (a, b) and (- a, b)
Answer: The slope is found using the two-point formula:
\( m = \frac{y_2 - y_1}{x_2 - x_1} \Rightarrow \frac{b - b}{-a - a} = 0 \)
Since \( m = 0 \), this represents a horizontal line.
Using the two-point form of a line equation:
\( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \)
\( y - b = 0(x - a) \)
\( y = b \)
The required equation of the line is \( y = b \).
In simple words: When two points share the same y-coordinate, the line connecting them is horizontal and runs parallel to the x-axis. Its equation is simply the y-value itself.

Exam Tip: Recognize that equal y-coordinates mean zero slope - always resulting in a horizontal line of the form \( y = c \).

Question 16. Find the angle which the line joining the points \( (\sqrt{3}, \sqrt{6}) \) and \( (\sqrt{2}, \sqrt{3}) \) makes with the x - axis.
Answer: To find the angle, first compute the slope using the two-point formula:
\( m = \frac{y_2 - y_1}{x_2 - x_1} \Rightarrow \frac{\sqrt{6} - \sqrt{3}}{\sqrt{2} - 1} = \frac{\sqrt{3}(\sqrt{2} - 1)}{(\sqrt{2} - 1)} = \sqrt{3} \)
Since the slope equals \( \tan\theta \), we have \( \tan\theta = \sqrt{3} \), which gives \( \theta = 60° \).
The angle the line makes with the positive x-axis is \( 60° \).
In simple words: Calculate the slope by dividing the change in y by the change in x. The slope equals the tangent of the angle. Since tangent of 60° is the square root of 3, the angle is 60°.

Exam Tip: Remember that \( m = \tan\theta \) links slope directly to angle - always simplify the slope first to match a standard trig value.

Question 17. Prove that the points A(1, 4), B(3, - 2) and C(4, - 5) are collinear. Also, find the equation of the line on which these points lie.
Answer: Three points lie on the same line - they are collinear - if the slope between the first two points equals the slope between the second two points.
Slope of AB:
\( m_{AB} = \frac{-2 - 4}{3 - 1} = \frac{-6}{2} = -3 \)
Slope of BC:
\( m_{BC} = \frac{-5 - (-2)}{4 - 3} = \frac{-3}{1} = -3 \)
Since \( m_{AB} = m_{BC} = -3 \), the points are verified to be collinear.
Using the two-point form with point A(1, 4):
\( y - 4 = -3(x - 1) \)
\( y - 4 + 3x - 3 = 0 \)
\( 3x + y - 7 = 0 \)
The equation of the line is \( 3x + y - 7 = 0 \).
In simple words: Check if three points sit on one line by testing whether the slope between points 1-2 matches the slope between points 2-3. If the slopes are equal, they are collinear. Then write the line equation using any one point and the shared slope.

Exam Tip: Always verify collinearity by comparing slopes - if slopes match exactly, the points are definitely collinear. Use the simplest point for the equation to avoid arithmetic errors.

Question 18. If A(0, 0), B(2, 4) and C(6, 4) are the vertices of a \( \triangle ABC \), find the equations of its sides.
Answer: Since vertex A is at the origin, lines through A take the form \( y = mx \).
For side AB, the slope is:
\( m = \frac{4 - 0}{2 - 0} = \frac{4}{2} = 2 \)
So the equation of line AB is \( y = 2x \).
For side AC, the slope is:
\( m = \frac{4 - 0}{6 - 0} = \frac{4}{6} = \frac{2}{3} \)
Using \( y = mx \): \( y = \frac{2}{3}x \Rightarrow 2x - 3y = 0 \)
So the equation of line AC is \( 2x - 3y = 0 \).
For side BC, both B(2, 4) and C(6, 4) have the same y-coordinate, indicating a horizontal line parallel to the x-axis:
\( y = 4 \)
The required equations are:
AB: \( y = 2x \)
AC: \( 2x - 3y = 0 \)
BC: \( y = 4 \)
In simple words: For a side of a triangle, compute its slope using the two endpoints. A line through the origin becomes \( y = mx \). If two points have matching y-coordinates, the line is horizontal. Convert slope-intercept form to standard form when needed.

Exam Tip: Identify special cases: when one vertex is the origin, lines become \( y = mx \); when two points share a y-coordinate, the line is always horizontal.

Question 19. If A(- 1, 6), B(- 3, - 9) and C(5, - 8) are the vertices of a \( \triangle ABC \), find the equations of its medians.
Answer: A median connects a vertex to the midpoint of the opposite side. First, find the midpoints L, M, N of sides BC, AC, AB respectively using the midpoint formula:
\( (x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
Midpoint L of BC:
\( L = \left( \frac{-3 + 5}{2}, \frac{-9 + (-8)}{2} \right) = \left( 1, \frac{-17}{2} \right) \)
Midpoint M of AC:
\( M = \left( \frac{-1 + 5}{2}, \frac{6 + (-8)}{2} \right) = (2, -1) \)
Midpoint N of AB:
\( N = \left( \frac{-1 + (-3)}{2}, \frac{6 + (-9)}{2} \right) = \left( -2, \frac{-3}{2} \right) \)
Now find the equations of medians AL, BM, CN using the two-point form.
For median AL passing through A(- 1, 6) and L(1, -17/2):
\( y - 6 = \frac{-17/2 - 6}{1 - (-1)}(x - (-1)) = \frac{-29/2}{2}(x + 1) = \frac{-29}{4}(x + 1) \)
\( 4(y - 6) = -29(x + 1) \)
\( 4y - 24 + 29x + 29 = 0 \)
\( 29x + 4y + 5 = 0 \)
For median BM passing through B(- 3, - 9) and M(2, -1):
\( y - (-9) = \frac{-1 - (-9)}{2 - (-3)}(x - (-3)) = \frac{8}{5}(x + 3) \)
\( 5(y + 9) = 8(x + 3) \)
\( 5y + 45 = 8x + 24 \)
\( 8x - 5y - 21 = 0 \)
For median CN passing through C(5, - 8) and N(-2, -3/2):
\( y - (-8) = \frac{-3/2 - (-8)}{-2 - 5}(x - 5) = \frac{13/2}{-7}(x - 5) = \frac{13}{-14}(x - 5) \)
\( 4(y + 8) = 13(x - 5) \)
\( 4y + 32 = 13x - 65 \)
\( 13x - 4y - 97 = 0 \)
The required equations of the medians are:
AL: \( 29x + y + 5 = 0 \)
BM: \( 8x - 5y - 21 = 0 \)
CN: \( 13x - 4y - 97 = 0 \)
In simple words: Find the midpoint of each side using the midpoint formula. Then use the two-point form to get the equation of the line joining a vertex to the opposite side's midpoint. This line is a median of the triangle.

Exam Tip: Label midpoints clearly and double-check midpoint calculations - a single arithmetic error here cascades into wrong median equations.

Question 20. Find the equation of the perpendicular bisector of the line segment whose end points are A(10, 4) and B(- 4, 9).
Answer: A perpendicular bisector is a line that is perpendicular to a given line segment and passes through its midpoint.
First, find the midpoint M of segment AB:
\( M = \left( \frac{10 + (-4)}{2}, \frac{4 + 9}{2} \right) = \left( 3, \frac{13}{2} \right) \)
Next, find the slope of line AB:
\( m_1 = \frac{9 - 4}{-4 - 10} = \frac{5}{-14} = -\frac{5}{14} \)
Since perpendicular lines satisfy \( m_1 \cdot m_2 = -1 \), the slope of the perpendicular bisector is:
\( m_2 = -\frac{1}{m_1} = -\frac{1}{-5/14} = \frac{14}{5} \)
Using the two-point form with M and slope \( m_2 \):
\( y - \frac{13}{2} = \frac{14}{5}(x - 3) \)
\( 5\left(y - \frac{13}{2}\right) = 14(x - 3) \)
\( 5y - \frac{65}{2} = 14x - 42 \)
\( 10y - 65 = 28x - 84 \)
\( 28x - 10y - 19 = 0 \)
The equation of the perpendicular bisector is \( 28x - 10y - 19 = 0 \).
In simple words: Find the midpoint of the segment. Calculate the slope of the original segment, then find the negative reciprocal to get the perpendicular slope. Write the line equation passing through the midpoint with the perpendicular slope.

Exam Tip: The key relationship is \( m_1 \times m_2 = -1 \) for perpendicular lines - use this to flip and negate the slope. Always verify your midpoint calculation first.

Question 22. If A(4, 3), B(0, 0) and C(2, 3) are the vertices of a \( \triangle ABC \), find the equation of the bisector of \( \angle A \).
Answer: An angle bisector from vertex A divides the angle into two equal parts. To find its equation, use the angle-bisector theorem: if AD bisects angle A, then the angle between sides AB and AD equals the angle between sides AC and AD.
First, find the slopes of sides AB and AC:
\( m_{AB} = \frac{0 - 3}{0 - 4} = \frac{-3}{-4} = \frac{3}{4} \)
\( m_{AC} = \frac{3 - 3}{2 - 4} = \frac{0}{-2} = 0 \)
Let m be the slope of the angle bisector AD. Using the angle formula between two lines:
\( \tan\theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \)
The angle between AB (slope 3/4) and AD (slope m) equals the angle between AD (slope m) and AC (slope 0):
\( \left| \frac{m - 3/4}{1 + m \cdot 3/4} \right| = \left| \frac{0 - m}{1 + 0 \cdot m} \right| \)
\( \left| \frac{m - 3/4}{1 + 3m/4} \right| = |{-m}| \)
\( \left| \frac{4m - 3}{4 + 3m} \right| = |m| \)
Taking the case: \( \frac{4m - 3}{4 + 3m} = m \)
\( 4m - 3 = m(4 + 3m) \)
\( 4m - 3 = 4m + 3m^2 \)
\( -3 = 3m^2 \)
This gives no real solution. Try: \( \frac{4m - 3}{4 + 3m} = -m \)
\( 4m - 3 = -m(4 + 3m) \)
\( 4m - 3 = -4m - 3m^2 \)
\( 3m^2 + 8m - 3 = 0 \)
Using the quadratic formula: \( m = \frac{-8 \pm \sqrt{64 + 36}}{6} = \frac{-8 \pm 10}{6} \)
\( m = \frac{2}{6} = \frac{1}{3} \) or \( m = \frac{-18}{6} = -3 \)
Since we need the interior angle bisector, \( m = \frac{1}{3} \) is the correct slope.
Now find the equation passing through A(4, 3) with slope 1/3:
\( y = mx + c \)
\( 3 = \frac{1}{3}(4) + c \)
\( 3 = \frac{4}{3} + c \)
\( c = 3 - \frac{4}{3} = \frac{9 - 4}{3} = \frac{5}{3} \)
\( y = \frac{1}{3}x + \frac{5}{3} \)
\( 3y = x + 5 \)
\( x - 3y + 5 = 0 \)
The equation of the angle bisector of \( \angle A \) is \( x - 3y + 5 = 0 \).
In simple words: An angle bisector divides an angle equally. Find the slopes of the two sides meeting at that angle. Use the angle formula to set up an equation: the angle from one side to the bisector equals the angle from the bisector to the other side. Solve for the bisector's slope, then write its equation through the vertex.

Exam Tip: The angle-bisector formula can be tricky - set up the absolute-value equation carefully and consider both cases. The interior bisector is typically the one with a slope between the two side slopes (or matching the geometry of the figure).

Question 23. The midpoints of the sides BC, CA and AB of a \( \triangle ABC \) are D(2, 1), E(- 5, 7) and F(- 5, - 5) respectively. Find the equations of the sides of \( \triangle ABC \).
Answer: Let the vertices of the triangle be A(a, b), B(c, d), C(e, f). Using the midpoint formula:
For side BC with midpoint D(2, 1):
\( 2 = \frac{c + e}{2}, \quad 1 = \frac{d + f}{2} \)
\( c + e = 4, \quad d + f = 2 \quad \text{...(i)} \)
For side CA with midpoint E(- 5, 7):
\( -5 = \frac{a + e}{2}, \quad 7 = \frac{b + f}{2} \)
\( a + e = -10, \quad b + f = 14 \quad \text{...(ii)} \)
For side AB with midpoint F(- 5, - 5):
\( -5 = \frac{a + c}{2}, \quad -5 = \frac{b + d}{2} \)
\( a + c = -10, \quad b + d = -10 \quad \text{...(iii)} \)
Subtracting (i) from (ii):
\( a - c = -14, \quad b - d = 12 \quad \text{...(iv)} \)
Adding (iii) and (iv):
\( 2a = -24 \Rightarrow a = -12 \)
\( 2b = 2 \Rightarrow b = 1 \)
Substituting into (iii):
\( c = -10 - a = -10 - (-12) = 2 \)
\( d = -10 - b = -10 - 1 = -11 \)
Substituting into (i):
\( e = 4 - c = 4 - 2 = 2 \)
\( f = 2 - d = 2 - (-11) = 13 \)
So the vertices are A(- 12, 1), B(2, - 11), C(2, 13).
Now find the equations of the sides using the two-point form:
For side AB:
\( y - 1 = \frac{-11 - 1}{2 - (-12)}(x - (-12)) = \frac{-12}{14}(x + 12) \)
\( 14(y - 1) = -12(x + 12) \)
\( 14y - 14 + 12x + 144 = 0 \)
\( 12x + 14y + 130 = 0 \)
\( 6x + 7y + 65 = 0 \)

For side CA:
\( y - 13 = \frac{1 - 13}{-12 - 2}(x - 2) = \frac{-12}{-14}(x - 2) = \frac{6}{7}(x - 2) \)
\( 7(y - 13) = 6(x - 2) \)
\( 7y - 91 = 6x - 12 \)
\( 6x - 7y + 79 = 0 \)
The required equations of the sides are:
AB: \( 6x + 7y + 65 = 0 \)
BC: \( x = 2 \)
CA: \( 6x - 7y + 79 = 0 \)
In simple words: Use the three midpoint equations to set up a system. Solve for the three vertex coordinates by combining the equations methodically. Once you have the vertices, apply the two-point form to write each side's equation.

Exam Tip: Keep the three midpoint equations organized - label them clearly and solve the system step-by-step. Watch for vertical or horizontal lines where two vertices share a coordinate.

Question 24. If A(1, 4), B(2, - 3) and C( - 1, - 2) are the vertices of a ΔABC, find the equation of (i) the median through A (ii) the altitude through A (iii) the perpendicular bisector of BC
Answer: (i) To get the equation of median AD, first locate the midpoint of side BC.

For side BC (midpoint D): \( (x, y) = \left(\frac{2 + (-1)}{2}, \frac{-3 + (-2)}{2}\right) \)

\( (x, y) = \left(\frac{1}{2}, \frac{-5}{2}\right) \)

Using the two-point form of a line equation:

\( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \)

\( y - 4 = \frac{\frac{-5}{2} - 4}{\frac{1}{2} - 1}(x - 1) \)

\( y - 4 = \frac{\frac{-5 - 8}{2}}{\frac{1 - 2}{2}}(x - 1) \)

\( y - 4 = \frac{-13}{-1}(x - 1) \)

\( y - 4 = 13(x - 1) \)

\( y - 4 = 13x - 13 \)

\( 13x - y - 9 = 0 \)

(ii) To find the equation of the altitude through A, first determine the slope of side BC, then apply the relationship for perpendicular lines where \( m_1 \cdot m_2 = -1 \).

Slope of BC: \( m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - (-3)}{-1 - 2} = \frac{-2 + (-3)}{-3} = \frac{-5}{-3} = \frac{5}{3} \)

Slope of altitude through A: \( m_1 \cdot m_1' = -1 \) gives \( \frac{5}{3} \cdot m_1' = -1 \), so \( m_1' = \frac{-3}{5} \)

Using slope-intercept form \( y = mx + c \), substitute point A(1, 4):

\( 4 = \frac{-3}{5}(1) + c \)

\( c = 4 + \frac{3}{5} = \frac{20 + 3}{5} = \frac{23}{5} \)

The equation becomes: \( y = \frac{-3}{5}x + \frac{23}{5} \)

\( 5y = -3x + 23 \)

\( 3x + 5y - 23 = 0 \)

(iii) For the perpendicular bisector, use the slope found in part (ii) and the midpoint D from part (i). Since the perpendicular bisector passes through the midpoint of BC and has slope \( \frac{-3}{5} \):

Using \( y = mx + c \) with point \( D\left(\frac{1}{2}, \frac{-5}{2}\right) \):

\( \frac{-5}{2} = \frac{-3}{5} \cdot \frac{1}{2} + c \)

\( \frac{-5}{2} = \frac{-3}{10} + c \)

\( c = \frac{-5}{2} + \frac{3}{10} = \frac{-25 + 3}{10} = \frac{-22}{10} = \frac{-11}{5} \)

The equation becomes: \( y = \frac{-3}{5}x + \frac{-11}{5} \)

\( 5y = -3x - 11 \)

\( 3x + y + 11 = 0 \)
In simple words: A median links a vertex to the centre point of the opposite side. An altitude drops perpendicularly from a vertex to the opposite side. A perpendicular bisector cuts a line segment at its middle point at a right angle.

Exam Tip: Always compute the midpoint first when working with medians or perpendicular bisectors. Remember that perpendicular slopes satisfy \( m_1 \cdot m_2 = -1 \).

Exercise 20D

Question 1. Find the equation of the line whose (i) slope = 3 and y - intercept = 5 (ii) slope = - 1 and y - intercept = 4 (iii) slope = - \( \frac{2}{5} \) and y - intercept = - 3
Answer: (i) The formula to use is \( y = mx + c \), where m is the slope and c represents the y - intercept.

Here, m = 3 and c = 5.

So, \( y = (3)x + (5) \)

\( y = 3x + 5 \)

(ii) The formula to use is \( y = mx + c \), where m is the slope and c represents the y - intercept.

Here, m = - 1 and c = 4.

So, \( y = ( - 1)x + (4) \)

\( x + y = 4 \)

(iii) The formula to use is \( y = mx + c \), where m is the slope and c represents the y - intercept.

Here, \( m = - \frac{2}{5} \) and c = - 3.

So, \( y = \left( - \frac{2}{5}\right)x + ( - 3) \)

\( 5y = - 2x - 15 \)

\( 2x + 5y + 15 = 0 \)
In simple words: Once you know the slope and where the line meets the y - axis, plug those values into the basic line formula. This gives you the equation right away.

Exam Tip: Memorize \( y = mx + c \) — it is the fastest route to a line equation when both slope and intercept are given.

Question 2. Find the equation of the line which makes an angle of 30° with the positive direction of the x - axis and cuts off an intercept of 4 units with the negative direction of the y - axis.
Answer: The line makes an angle of 30° with the x - axis, so the y - intercept equals - 4.

Using the formula \( y = mx + c \), where m is the slope and c is the y - intercept:

The slope \( m = \tan(30°) = \frac{1}{\sqrt{3}} \)

The equation of the line is \( y = \frac{1}{\sqrt{3}}x - 4 \)

\( \sqrt{3}y = x - 4\sqrt{3} \)

\( x - \sqrt{3}y = 4\sqrt{3} \)
In simple words: When a line tilts at a specific angle to the horizontal, its slope comes from the tangent of that angle. Add the y - intercept value, and you get the full equation.

Exam Tip: For angle-based problems, always convert the angle to slope using \( m = \tan(\theta) \) before writing the final equation.

Question 3. Find the equation of the line whose inclination is \( \frac{5\pi}{6} \) and which makes an intercept of 6 units on the negative direction of the y - axis.
Answer: Given that the inclination angle is \( \theta = \frac{5\pi}{6} \) and the y - intercept is - 6.

The slope is \( m = \tan(\theta) = \tan\left(\frac{5\pi}{6}\right) = -\frac{1}{\sqrt{3}} \)

The y - intercept equals - 6 (six units in the negative direction).

Using the formula \( y = mx + c \):

The equation of the line is \( y = -\frac{1}{\sqrt{3}}x - 6 \)

\( \sqrt{3}y = -x - 6\sqrt{3} \)

\( x + \sqrt{3}y + 6\sqrt{3} = 0 \)
In simple words: An inclination angle in radians must be converted to its tangent value to get the slope. Then use the standard form with the intercept value.

Exam Tip: When the inclination angle is given in radians (especially multiples of \( \pi \)), carefully apply the tangent function to find the slope — watch for negative slopes in the second quadrant.

Question 4. Find the equation of the line cutting off an intercept - 2 from the y - axis and equally inclined to the axes.
Answer: The line is equally inclined to both axes, meaning it makes the same angle with each axis. Since the coordinate axes meet at 90°, if each inclination angle is θ, then \( \theta + \theta = 90° \), giving \( \theta = 45° \).

The slope of the line is \( m = \tan(45°) = 1 \).

The y - intercept equals - 2.

Using the formula \( y = mx + c \):

The equation of the line is \( y = 1 \cdot x + (-2) = x - 2 \)

\( x - y = 2 \)
In simple words: A line equally inclined to both axes makes a 45° angle with each one. This always gives a slope of 1 or - 1, depending on direction.

Exam Tip: Equal inclination to the axes is a quick signal: the slope must be ±1. Use the intercept to determine the full equation immediately.

Question 5. Find the equation of the bisectors of the angles between the coordinate axes.
Answer: The coordinate axes are the lines \( x = 0 \) (the y - axis) and \( y = 0 \) (the x - axis). Using the angle bisector formula, if two lines are \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), their angle bisectors are given by:

\( \frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm\frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}} \)

For the axes (x = 0 and y = 0):

\( \frac{x}{\sqrt{1^2}} = \pm\frac{y}{\sqrt{1^2}} \)

\( x = \pm y \)

The two angle bisectors are \( x = y \) and \( x = -y \).
In simple words: The lines that split the four right angles at the origin into equal parts are the diagonals: one going up-right (\( x = y \)) and one going up-left (\( x = -y \)).

Exam Tip: The angle bisectors of the coordinate axes always produce the simplest equations: \( x = y \) and \( x = -y \). Recognize this pattern instantly.

Question 6. Find the equation of the line through the point ( - 1, 5) and making an intercept of - 2 on the y - axis.
Answer: The y - intercept is - 2, and the line passes through ( - 1, 5).

Using the formula \( y = mx + c \), where c = - 2:

\( y = mx - 2 \)

Substitute the point ( - 1, 5) into the equation:

\( 5 = m(-1) - 2 \)

\( 5 = -m - 2 \)

\( m = -(5 + 2) = -7 \)

The equation becomes \( y = -7x - 2 \)

\( 7x + y + 2 = 0 \)
In simple words: When you know the intercept and one point on the line, substitute the point to find the slope. Then write the full equation using both pieces of information.

Exam Tip: Always substitute the given point into the equation with the known intercept to solve for the unknown slope — this method is fast and reliable.

Question 7. Find the equation of the line which is parallel to the line 2x - 3y = 8 and whose y - intercept is 5 units.
Answer: A line parallel to \( 2x - 3y = 8 \) has the same slope. Both parallel lines can be written as \( 2x - 3y = k \) for different values of k.

Rearrange to slope-intercept form: \( 3y = 2x - k \), or \( y = \frac{2}{3}x - \frac{k}{3} \).

The y - intercept is \( -\frac{k}{3} = 5 \), so \( k = -15 \).

The equation of the required line is \( 2x - 3y = -15 \)

\( 2x - 3y + 15 = 0 \)
In simple words: Parallel lines share the same slope. Keep the same coefficients for x and y, then adjust the constant to match the desired intercept.

Exam Tip: For parallel lines, match the slope first by keeping the x and y coefficients identical, then solve for the constant using the given intercept.

Question 8. Find the equation of the line passing through the point (0, 3) and perpendicular to the line x - 2y + 5 = 0
Answer: The given line is \( x - 2y + 5 = 0 \). Rewrite as \( 2y = x + 5 \), or \( y = \frac{1}{2}x + \frac{5}{2} \).

The slope of this line is \( \frac{1}{2} \).

For perpendicular lines, \( m_1 \cdot m_2 = -1 \), so the slope of the perpendicular line is \( m = -2 \).

Using \( y = mx + c \) with the point (0, 3):

\( 3 = (-2)(0) + c \), so \( c = 3 \).

The required equation is \( y = -2x + 3 \)

\( 2x + y = 3 \)
In simple words: Extract the slope of the original line. Take its negative reciprocal to find the perpendicular slope. Then use the given point to find the intercept.

Exam Tip: For perpendicularity, always multiply the slopes: if their product is - 1, the lines are perpendicular. This check takes seconds and prevents errors.

Question 9. Find the equation of the line passing through the point (2, 3) and perpendicular to the line 4x + 3y = 10
Answer: Rewrite \( 4x + 3y = 10 \) in slope-intercept form: \( 3y = -4x + 10 \), or \( y = -\frac{4}{3}x + \frac{10}{3} \).

The slope of this line is \( -\frac{4}{3} \).

For perpendicular lines, \( m_1 \cdot m_2 = -1 \), so the slope of the perpendicular line is \( m = \frac{3}{4} \).

Using \( y = mx + c \) with the point (2, 3):

\( 3 = \frac{3}{4}(2) + c \)

\( 3 = \frac{3}{2} + c \)

\( c = 3 - \frac{3}{2} = \frac{6 - 3}{2} = \frac{3}{2} \)

The required equation is \( y = \frac{3}{4}x + \frac{3}{2} \)

\( 4y = 3x + 6 \)

\( 3x - 4y + 6 = 0 \)
In simple words: Find the slope of the given line, flip it upside down and change its sign to get the perpendicular slope. Apply the point to determine the intercept.

Exam Tip: The perpendicular slope is the negative reciprocal: if the original slope is \( -\frac{4}{3} \), the perpendicular slope is \( \frac{3}{4} \). Double-check this relationship before finalizing.

Question 10. Find the equation of the line passing through the point (2, 4) and perpendicular to the x - axis.
Answer: A line perpendicular to the x - axis (which is horizontal) is vertical. Vertical lines have the form \( x = c \), where c is a constant.

Since the line passes through (2, 4), the value of c equals the x - coordinate of the point.

\( c = 2 \)

The required equation is \( x = 2 \)
In simple words: Any line perpendicular to the horizontal x - axis must be vertical. A vertical line through a point simply uses that point's x - coordinate as its equation.

Exam Tip: Perpendicular to the x - axis always means a vertical line. Perpendicular to the y - axis means a horizontal line. Recognize these instantly.

Question 11. Find the equation of the line that has x - intercept - 3 and which is perpendicular to the line 3x + 5y = 4
Answer: Rewrite \( 3x + 5y = 4 \) in slope-intercept form: \( 5y = -3x + 4 \), or \( y = -\frac{3}{5}x + \frac{4}{5} \).

The slope of this line is \( -\frac{3}{5} \).

For perpendicular lines, the slope of the perpendicular line is \( m = \frac{5}{3} \).

Using the form \( y = mx + c \):

\( y = \frac{5}{3}x + c \)

The x - intercept occurs when \( y = 0 \):

\( 0 = \frac{5}{3}x + c \), or \( x = -\frac{3c}{5} \)

Given that the x - intercept is - 3:

\( -3 = -\frac{3c}{5} \)

\( c = 5 \)

The required equation is \( y = \frac{5}{3}x + 5 \)

\( 3y = 5x + 15 \)

\( 5x - 3y + 15 = 0 \)
In simple words: Find the perpendicular slope, then use the x - intercept value to calculate the constant term. The x - intercept tells you where the line crosses the horizontal axis.

Exam Tip: For x - intercept problems, set \( y = 0 \) in your slope-intercept form. This quickly links the intercept value to the unknown constant.

Question 12. Find the equation of the line which is perpendicular to the line 3x + 2y = 8 and passes through the midpoint of the line joining the points (6, 4) and (4, - 2).
Answer: First, find the midpoint of the segment joining (6, 4) and (4, - 2):

Midpoint \( = \left(\frac{6 + 4}{2}, \frac{4 + (-2)}{2}\right) = \left(\frac{10}{2}, \frac{2}{2}\right) = (5, 1) \)

Next, find the slope of the perpendicular line. The given line is \( 3x + 2y = 8 \), or \( 2y = -3x + 8 \), or \( y = -\frac{3}{2}x + 4 \).

The slope is \( -\frac{3}{2} \). For perpendicular lines, the slope is \( m = \frac{2}{3} \).

Using \( y = mx + c \) with the point (5, 1):

\( 1 = \frac{2}{3}(5) + c \)

\( 1 = \frac{10}{3} + c \)

\( c = 1 - \frac{10}{3} = \frac{3 - 10}{3} = -\frac{7}{3} \)

The required equation is \( y = \frac{2}{3}x - \frac{7}{3} \)

\( 3y = 2x - 7 \)

\( 2x - 3y - 7 = 0 \)
In simple words: Find the midpoint first using the average of the coordinates. Then treat this midpoint as your known point and apply the perpendicular slope condition to get the final equation.

Exam Tip: Two-step problems like this one benefit from organizing your work: (1) find the midpoint or other geometric feature, (2) apply the slope or perpendicularity condition. Separate these steps clearly.

Question 13. Find the equation of the line whose y - intercept is - 3 and which is perpendicular to the line joining the points ( - 2, 3) and (4, - 5).
Answer: The slope of the line passing through ( - 2, 3) and (4, - 5) can be found using the two-point formula. We get slope = \( -\frac{4}{3} \). For a perpendicular line, the slope is the negative reciprocal, which gives us \( \frac{3}{4} \). Since the required line has y-intercept = - 3, its equation in slope-intercept form is \( y = \frac{3}{4}x - 3 \). Converting to standard form: \( 3x - 4y = 12 \).
In simple words: First find the slope of the given line. The perpendicular line has the opposite reciprocal slope. Use the y-intercept to write the final equation.

Exam Tip: Remember that slopes of perpendicular lines multiply to give - 1. Always verify your final equation by checking the y-intercept value.

Question 14. Find the equation of the line passing through ( - 3, 5) and perpendicular to the line through the points (2, 5) and ( - 3, 6).
Answer: Using the two-point form, the slope of the line joining (2, 5) and ( - 3, 6) is found to be \( -\frac{1}{5} \). The perpendicular line must have slope = 5 (the negative reciprocal). Since this line passes through ( - 3, 5), we can substitute into \( y = 5x + c \) to get: \( 5 = 5(-3) + c \), so \( c = 20 \). Therefore the equation is \( y = 5x + 20 \) or \( 5x - y + 20 = 0 \).
In simple words: Find the slope of the original line. The perpendicular slope is its opposite reciprocal. Use the given point to find the y-intercept and write the equation.

Exam Tip: Always double-check by substituting the given point into your final equation to confirm it lies on the line.

Question 15. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : 2. Find the equation of the line.
Answer: First, find the point that divides the segment joining (1, 0) and (2, 3) in the ratio 1:2 using the section formula. This point is \( \left(\frac{4}{3}, 1\right) \). The slope of the line segment is 3, so the perpendicular line has slope \( -\frac{1}{3} \). Using the point-slope form with the point \( \left(\frac{4}{3}, 1\right) \): \( 1 = -\frac{1}{3} \cdot \frac{4}{3} + c \), which gives \( c = \frac{13}{9} \). The equation is \( y = -\frac{1}{3}x + \frac{13}{9} \) or \( 3x + 9y = 13 \).
In simple words: Use the section formula to locate the dividing point. Find the perpendicular slope. Apply the point and slope to determine the final equation.

Exam Tip: The section formula with ratio m:n is crucial here. Verify your division point satisfies the ratio condition.

Exercise 20E

Question 1. Find the equation of the line which cuts off intercepts -3 and 5 on the x-axis and y-axis respectively.
Answer: Using the intercept form of a line, where \( \frac{x}{a} + \frac{y}{b} = 1 \) with a = -3 and b = 5, we get: \( \frac{x}{-3} + \frac{y}{5} = 1 \). Multiplying through to clear fractions: \( 5x - 3y = -15 \) or \( 5x - 3y + 15 = 0 \).
In simple words: The intercept form directly gives us the equation when we know where the line crosses the axes. Substitute the intercepts and simplify.

Exam Tip: Always identify whether the given numbers are x-intercepts or y-intercepts. The intercept form is quick for these standard problems.

Question 2. Find the equation of the line which cuts off intercepts 4 and -6 on the x-axis and y-axis respectively.
Answer: Using the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \) with a = 4 and b = -6: \( \frac{x}{4} + \frac{y}{-6} = 1 \). Simplifying: \( -3x + 2y = -12 \) or \( 3x - 2y - 12 = 0 \).
In simple words: Apply the intercept form with the given x and y intercepts. Clear fractions to get the standard form.

Exam Tip: Pay attention to the signs of the intercepts - negative intercepts change the position of the line significantly.

Question 3. Find the equation of the line and cuts off equal intercepts on the coordinate axes and passes through the point (4,7).
Answer: Since the intercepts are equal, let \( a = b \). Using the intercept form \( \frac{x}{a} + \frac{y}{a} = 1 \) or \( x + y = a \). Since the line passes through (4, 7): \( 4 + 7 = a \), so \( a = 11 \). The equation is \( x + y = 11 \).
In simple words: When intercepts are equal, the equation simplifies to x + y = a. Just substitute the given point to find a.

Exam Tip: Equal intercepts lead to a line with equation x + y = constant. This is much faster than the general intercept form.

Question 4. Find the equation of the line which passes through the point (3, -5) and cuts off intercepts on the axes which are equal in magnitude but opposite in sign.
Answer: When intercepts are equal in magnitude but opposite in sign, we have \( a = -b \). Using the intercept form: \( \frac{x}{a} + \frac{y}{-a} = 1 \) or \( x - y = a \). Since (3, -5) lies on the line: \( 3 - (-5) = a \), giving \( a = 8 \). Therefore the equation is \( x - y = 8 \).
In simple words: Opposite intercepts mean the line has the form x - y = constant. Plug in the given point to find that constant.

Exam Tip: Recognize when intercepts are opposite - it yields the simple form x - y = a, which is easier to work with than general intercept form.

Question 5. Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes, whose sum is 9.
Answer: Let the intercepts be a and b with \( a + b = 9 \), so \( b = 9 - a \). Using intercept form: \( \frac{x}{a} + \frac{y}{9-a} = 1 \). Since (2, 2) lies on the line: \( \frac{2}{a} + \frac{2}{9-a} = 1 \). Solving: \( 2(9-a) + 2a = a(9-a) \), which simplifies to \( a^2 - 9a + 18 = 0 \). Factoring: \( (a-3)(a-6) = 0 \), giving a = 3 or a = 6. For a = 3, b = 6: equation is \( 2x + y = 6 \). For a = 6, b = 3: equation is \( x + 2y = 6 \).
In simple words: Use the constraint that intercepts sum to 9. Express one intercept in terms of the other. Substitute the point to get a quadratic. Both solutions give valid line equations.

Exam Tip: Always check both solutions from the quadratic - both typically represent valid geometric configurations satisfying the problem conditions.

Question 6. Find the equation of the line which passes through the point (22, -6) and whose intercept on the x-axis exceeds the intercept on the y-axis by 5.
Answer: Let the y-intercept be b, so the x-intercept is \( a = b + 5 \). Using intercept form: \( \frac{x}{b+5} + \frac{y}{b} = 1 \). Since (22, -6) lies on the line: \( \frac{22}{b+5} + \frac{-6}{b} = 1 \). Simplifying: \( 22b - 6(b+5) = b(b+5) \), which becomes \( b^2 - 11b + 30 = 0 \). Factoring: \( (b-5)(b-6) = 0 \). For b = 5, a = 10: equation is \( x + 2y = 10 \). For b = 6, a = 11: equation is \( 6x + 11y = 66 \).
In simple words: Express the x-intercept in terms of the y-intercept using the given relationship. Substitute the point into the intercept form and solve the resulting quadratic.

Exam Tip: When one intercept exceeds another by a fixed amount, set up the relationship carefully. Both solutions should be verified by checking that the point satisfies each equation.

Question 7. Find the equation of the line whose portion intercepted between the axes is bisected at the point (3, -2).
Answer: Let A(a, 0) and B(0, b) be the points where the line meets the axes. Since the portion AB is bisected at (3, -2), using the midpoint formula: \( \frac{a+0}{2} = 3 \) and \( \frac{0+b}{2} = -2 \). This gives a = 6 and b = -4. Using intercept form: \( \frac{x}{6} + \frac{y}{-4} = 1 \). Simplifying: \( -2x + 3y = -12 \) or \( 2x - 3y = 12 \).
In simple words: If a segment is bisected at a point, the midpoint formula tells us the coordinates of the axis intercepts. Then apply the intercept form directly.

Exam Tip: The midpoint of the intercept segment lies on the line itself. This constraint uniquely determines both intercepts.

Question 8. Find the equation of the line whose portion intercepted between the coordinate axes is divided at the point (5, 6) in the ratio 3 : 1.
Answer: Let A(a, 0) and B(0, b) be the intercept points. The point (5, 6) divides AB in ratio 3:1 using the section formula: \( 5 = \frac{3 \cdot 0 + 1 \cdot a}{3+1} = \frac{a}{4} \), so a = 20. Similarly, \( 6 = \frac{3 \cdot b + 1 \cdot 0}{3+1} = \frac{3b}{4} \), so b = 8. Using intercept form: \( \frac{x}{20} + \frac{y}{8} = 1 \). Simplifying: \( 2x + 5y = 40 \).
In simple words: Apply the section formula to find where each axis is cut. The dividing point's coordinates directly reveal the intercepts through the section formula.

Exam Tip: The section formula is key here. Be careful with the ratio direction - verify that the computed intercepts actually divide the segment in the stated ratio.

Question 9. A straight line passes through the point (5, -2) and the portion of the line intercepted between the axes is divided at this point in the ratio 2 - 3. Find the equation of the line.
Answer: Let the intercepts be represented by point A(a, 0) on the x-axis and point B(0, b) on the y-axis. The given point (5, -2) divides the line segment AB in the ratio 2 - 3. Using the section formula:

\( (x, y) = \left( \frac{my_1 + nx_1}{m + n}, \frac{my_2 + nx_2}{m + n} \right) \)

\( (5, -2) = \left( \frac{2 \cdot 0 + 3 \cdot a}{5}, \frac{2 \cdot b + 3 \cdot 0}{5} \right) = \left( \frac{3a}{5}, \frac{2b}{5} \right) \)

From this, we get:

\( \frac{3a}{5} = 5 \implies a = \frac{25}{3} \)

\( \frac{2b}{5} = -2 \implies b = -5 \)

The intercept form of a line is \( \frac{x}{a} + \frac{y}{b} = 1 \)

\( \frac{x}{\frac{25}{3}} + \frac{y}{-5} = 1 \)

\( \frac{3x}{25} - \frac{y}{5} = 1 \)

\( \frac{3x}{25} - \frac{5y}{25} = 1 \)

\( 3x - 5y = 25 \)

Hence, the required equation of the line is 3x - 5y = 25.

Exam Tip: Always use the section formula carefully when a point divides a line segment in a given ratio. Check your intercept values by substituting them back into the original condition.

Question 10. If the straight line \( \frac{x}{a} + \frac{y}{b} = 1 \) passes through the points (8, -9) and (12, -15), find the values of a and b.
Answer: Since the line \( \frac{x}{a} + \frac{y}{b} = 1 \) passes through both given points, we substitute each point into the equation to get two separate conditions.

For point (8, -9):

\( \frac{8}{a} + \frac{-9}{b} = 1 \)

\( 8b - 9a = ab \) — equation 2

For point (12, -15):

\( \frac{12}{a} + \frac{-15}{b} = 1 \)

\( 12b - 15a = ab \) — equation 3

Subtracting equation 2 from equation 3:

\( (12b - 15a) - (8b - 9a) = 0 \)

\( 4b - 6a = 0 \)

\( a = \frac{2b}{3} \)

Substituting this relationship into equation 2:

\( 8b - 9 \cdot \frac{2b}{3} = \frac{2b}{3} \cdot b \)

\( 8b - 6b = \frac{2b^2}{3} \)

\( 2b = \frac{2b^2}{3} \)

\( b = 3 \)

Therefore, \( a = \frac{2 \cdot 3}{3} = 2 \)

Hence the values of a and b are 2 and 3 respectively.

Exam Tip: When a line passes through two distinct points, substitute both points independently and solve the resulting system of equations simultaneously to find unknown parameters.

Exercise 20F

Question 1. Find the equation of the line for which p = 3 and α = 450
Answer: The normal form of a line is expressed as \( x \cos \alpha + y \sin \alpha = p \), where p is the perpendicular distance from the origin and α is the angle the perpendicular makes with the positive x-axis direction.

Substituting the given values:

\( x \cos 450 + y \sin 450 = 3 \)

Using angle reduction: \( \cos 450 = \cos(360 + 90) = \cos 90 = 0 \) and \( \sin 450 = \sin(360 + 90) = \sin 90 = 1 \)

\( x \cdot 0 + y \cdot 1 = 3 \)

\( y = 3 \)

Hence the required equation of the line is y = 3.

Exam Tip: Always reduce angles greater than 360° by subtracting multiples of 360° until the angle falls within the standard range, then apply trigonometric identities.

Question 2. Find the equation of the line for which p = 5 and α = 1350
Answer: Using the normal form \( x \cos \alpha + y \sin \alpha = p \):

\( x \cos 1350 + y \sin 1350 = 5 \)

Simplifying the angle: \( \cos 1350 = \cos(4 \times 360 - 90) = \cos(-90) = \cos 90 = 0 \) and \( \sin 1350 = \sin(4 \times 360 - 90) = \sin(-90) = -\sin 90 = -1 \)

\( x \cdot 0 + y \cdot (-1) = 5 \)

\( -y = 5 \)

\( y = -5 \)

Hence the required equation of the line is y = -5.

Exam Tip: Watch for negative sine values when angles exceed 270°. Double-check your angle reduction to avoid sign errors in the final equation.

Question 3. Find the equation of the line for which p = 8 and α = 1500
Answer: Using the normal form \( x \cos \alpha + y \sin \alpha = p \):

\( x \cos 1500 + y \sin 1500 = 8 \)

Angle reduction gives: \( \cos 1500 = \cos(4 \times 360 + 60) = \cos 60 = \frac{1}{2} \) and \( \sin 1500 = \sin(4 \times 360 + 60) = \sin 60 = \frac{\sqrt{3}}{2} \)

\( x \cdot \frac{1}{2} + y \cdot \frac{\sqrt{3}}{2} = 8 \)

Multiplying both sides by 2:

\( x + \sqrt{3} y = 16 \)

Hence the required equation of the line is x + √3 y = 16.

Exam Tip: When the angle reduction leads to standard trigonometric values (30°, 45°, 60°), use their exact forms to avoid decimal approximations and keep the answer precise.

Question 4. Find the equation of the line for which p = 3 and α = 2250
Answer: Using the normal form \( x \cos \alpha + y \sin \alpha = p \):

\( x \cos 2250 + y \sin 2250 = 3 \)

Reducing the angle: \( \cos 2250 = \cos(6 \times 360 + 90) = \cos 90 = 0 \) and \( \sin 2250 = \sin(6 \times 360 + 90) = \sin 90 = 1 \)

\( x \cdot 0 + y \cdot 1 = 3 \)

\( y = 3 \)

Hence the required equation of the line is y = 3.

Exam Tip: Recognize that angles differing by multiples of 360° have identical sine and cosine values. Use this periodicity to simplify calculations.

Question 5. Find the equation of the line for which p = 2 and α = 3000
Answer: Using the normal form \( x \cos \alpha + y \sin \alpha = p \):

\( x \cos 3000 + y \sin 3000 = 2 \)

Angle reduction: \( \cos 3000 = \cos(8 \times 360 + 120) = \cos 120 = \cos(180 - 60) = -\cos 60 = -\frac{1}{2} \) and \( \sin 3000 = \sin(8 \times 360 + 120) = \sin 120 = \sin(180 - 60) = \sin 60 = \frac{\sqrt{3}}{2} \)

\( x \cdot \left(-\frac{1}{2}\right) + y \cdot \frac{\sqrt{3}}{2} = 2 \)

Multiplying by 2:

\( -x + \sqrt{3} y = 4 \)

\( x - \sqrt{3} y = -4 \)

Or equivalently, x - √3 y = -4.

Exam Tip: Use the supplementary angle formulas when reducing angles in the second or third quadrant: cos(180 - θ) = -cos θ and sin(180 - θ) = sin θ.

Question 6. Find the equation of the line for which p = 4 and α = 1800
Answer: Using the normal form \( x \cos \alpha + y \sin \alpha = p \):

\( x \cos 1800 + y \sin 1800 = 4 \)

Since 1800° = 5 × 360°, the angle reduces to: \( \cos 1800 = \cos 0 = 1 \) and \( \sin 1800 = \sin 0 = 0 \)

\( x \cdot 1 + y \cdot 0 = 4 \)

\( x = 4 \)

Hence the required equation of the line is x = 4.

Exam Tip: When an angle is a multiple of 360°, it represents a full rotation and returns to the starting direction. Check if the reduced angle falls on an axis (0°, 90°, 180°, 270°) to simplify quickly.

Question 7. The length of the perpendicular segment from the origin to a line is 2 units and the inclination of this perpendicular is α such that \( \sin \alpha = \frac{1}{3} \) and α is acute. Find the equation of the line.
Answer: Given that \( p = 2 \) units and \( \sin \alpha = \frac{1}{3} \), we need to find \( \cos \alpha \).

Using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \):

\( \left(\frac{1}{3}\right)^2 + \cos^2 \alpha = 1 \)

\( \frac{1}{9} + \cos^2 \alpha = 1 \)

\( \cos^2 \alpha = \frac{8}{9} \)

Since α is acute, \( \cos \alpha = \frac{2\sqrt{2}}{3} \)

Using the normal form \( x \cos \alpha + y \sin \alpha = p \):

\( x \cdot \frac{2\sqrt{2}}{3} + y \cdot \frac{1}{3} = 2 \)

Multiplying by 3:

\( 2\sqrt{2} x + y = 6 \)

Hence the required equation of the line is 2√2 x + y = 6 or √8 x + y = 6.

Exam Tip: Always use the Pythagorean identity to find the missing trigonometric ratio. Remember to choose the correct sign based on whether the angle is acute, obtuse, or lies in a specific quadrant.

Question 8. Find the equation of the line which is at a distance of 3 units from the origin such that \( \tan \alpha = \frac{5}{12} \), where α is the acute angle which this perpendicular makes with the positive direction of the x-axis.
Answer: Given \( p = 3 \) units and \( \tan \alpha = \frac{5}{12} \).

From the right triangle with opposite side 5 and adjacent side 12, the hypotenuse is:

\( \text{hyp} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \)

Therefore:

\( \sin \alpha = \frac{5}{13} \) and \( \cos \alpha = \frac{12}{13} \)

Using the normal form \( x \cos \alpha + y \sin \alpha = p \):

\( x \cdot \frac{12}{13} + y \cdot \frac{5}{13} = 3 \)

Multiplying by 13:

\( 12x + 5y = 39 \)

Hence the required equation of the line is 12x + 5y = 39.

Exam Tip: When given the tangent of an angle, construct a right triangle to find the sine and cosine values. The hypotenuse is always the square root of the sum of squares of the two legs.

Exercise 20G

Question 1. Reduce the equation 2x - 3y - 5 = 0 to slope-intercept form, and find from it the slope and y-intercept.
Answer: Rearranging the given equation 2x - 3y - 5 = 0:

\( 2x - 5 = 3y \)

\( y = \frac{2}{3}x - \frac{5}{3} \)

This is now in the slope-intercept form \( y = mx + c \), where m is the slope and c is the y-intercept.

Comparing with the standard form:

\( m = \frac{2}{3} \) and \( c = -\frac{5}{3} \)

In simple words: Rearrange the equation to isolate y on one side. The coefficient of x is the slope, and the constant term is the y-intercept (the point where the line crosses the y-axis).

Exam Tip: Always isolate y to get the slope-intercept form. The slope tells you how steep the line is, while the y-intercept shows where it meets the y-axis.

Question 2. Reduce the equation 5x + 7y - 35 = 0 to slope-intercept form, and hence find the slope and the y-intercept of the line.
Answer: Rearranging the given equation 5x + 7y - 35 = 0:

\( 7y = 35 - 5x \)

\( 7y = -5x + 35 \)

\( y = -\frac{5}{7}x + 5 \)

This is in the slope-intercept form \( y = mx + c \).

Comparing:

\( m = -\frac{5}{7} \) and \( c = 5 \)

Therefore, the slope is \( -\frac{5}{7} \) and the y-intercept is 5.

In simple words: The line goes down from left to right (negative slope) and crosses the y-axis at the point (0, 5).

Exam Tip: A negative slope indicates the line descends as x increases. Always double-check your algebra when isolating y to ensure the signs are correct.

Question 3. Reduce the equation y + 5 = 0 to slope-intercept form, and hence find the slope and the y-intercept of the line.
Answer: Rearranging the equation y + 5 = 0:

\( y = -5 \)

This is in the slope-intercept form \( y = mx + c \).

Comparing with the standard form:

\( m = 0 \) and \( c = -5 \)

Therefore, the slope is 0 and the y-intercept is -5.

In simple words: This is a horizontal line that never rises or falls. It stays at height -5 throughout, crossing the y-axis at the point (0, -5).

Exam Tip: A horizontal line always has a slope of zero. The equation has only a y-value and no x-term, which is the signature of a horizontal line.

Question 4. Reduce the equation 3x - 4y + 12 = 0 to intercepts form. Hence, find the length of the portion of the line intercepted between the axes.
Answer: Rearranging the equation 3x - 4y + 12 = 0:

\( 3x - 4y = -12 \)

Dividing throughout by -12:

\( \frac{3x}{-12} - \frac{4y}{-12} = 1 \)

\( \frac{x}{-4} + \frac{y}{3} = 1 \)

This is in the intercepts form \( \frac{x}{a} + \frac{y}{b} = 1 \).

From this, the x-intercept is -4 and the y-intercept is 3. The line crosses the axes at points (-4, 0) and (0, 3).

Using the distance formula between two points:

\( d = \sqrt{(-4 - 0)^2 + (0 - 3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \)

Therefore, the length of the portion of the line intercepted between the axes is 5 units.

In simple words: Find where the line crosses each axis to get two points, then measure the straight-line distance between them using the distance formula.

Exam Tip: The distance between the intercepts on the two axes represents the length of the line segment visible between the axes. Always apply the distance formula correctly with both coordinate points.

Question 5. Reduce the equation 5x - 12y = 60 to intercepts form. Hence, find the length of the portion of the line intercepted between the axes.
Answer: Rearranging the equation 5x - 12y = 60:

Dividing throughout by 60:

\( \frac{5x}{60} - \frac{12y}{60} = 1 \)

\( \frac{x}{12} - \frac{y}{5} = 1 \)

\( \frac{x}{12} + \frac{y}{-5} = 1 \)

This is in the intercepts form \( \frac{x}{a} + \frac{y}{b} = 1 \).

From this, the x-intercept is 12 and the y-intercept is -5. The line crosses the axes at points (12, 0) and (0, -5).

Using the distance formula:

\( d = \sqrt{(12 - 0)^2 + (0 - (-5))^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \)

Therefore, the length of the portion of the line intercepted between the axes is 13 units.

In simple words: Find the two intercept points on the coordinate axes and calculate the distance between them using the Pythagorean theorem or the distance formula.

Exam Tip: When converting to intercepts form, divide the entire equation by the constant term on the right-hand side. This directly reveals the intercept values on both axes.

Question 6. Find the inclination of the line:
(i) \( x + \sqrt{3} y + 6 = 0 \)
(ii) \( 3x + 3y + 8 = 0 \)
(iii) \( \sqrt{3} x - y - 4 = 0 \)
Answer: The inclination of a line is the angle α that the line makes with the positive x-axis direction. We find it using the slope-intercept form \( y = mx + c \) where \( \tan \alpha = m \).

(i) For \( x + \sqrt{3} y + 6 = 0 \):

Rearranging: \( \sqrt{3} y = -x - 6 \)

\( y = -\frac{1}{\sqrt{3}}x - \frac{6}{\sqrt{3}} \)

The slope is \( m = -\frac{1}{\sqrt{3}} \)

\( \tan \alpha = -\frac{1}{\sqrt{3}} = -\tan 30° = \tan(180° - 30°) = \tan 150° \)

\( \alpha = 150° = \frac{5\pi}{6} \)

(ii) For \( 3x + 3y + 8 = 0 \):

Rearranging: \( 3y = -3x - 8 \)

\( y = -x - \frac{8}{3} \)

The slope is \( m = -1 \)

\( \tan \alpha = -1 = \tan(180° - 45°) = \tan 135° \)

\( \alpha = 135° = \frac{3\pi}{4} \)

(iii) For \( \sqrt{3} x - y - 4 = 0 \):

Rearranging: \( y = \sqrt{3}x - 4 \)

The slope is \( m = \sqrt{3} \)

\( \tan \alpha = \sqrt{3} = \tan 60° \)

\( \alpha = 60° = \frac{\pi}{3} \)

In simple words: Convert the equation to slope-intercept form to get the slope. Then use the inverse tangent function to find the angle the line makes with the horizontal. If the slope is negative, the angle will be obtuse (between 90° and 180°).

Exam Tip: Remember that \( \tan 30° = \frac{1}{\sqrt{3}} \), \( \tan 45° = 1 \), and \( \tan 60° = \sqrt{3} \). Use these standard values to quickly identify the inclination angle. For negative slopes, use supplementary angle relationships.

Question 7. Reduce the equation x + y - √2 = 0 to the normal form x cos α + y sin α = p, and hence find the values of α and p.
Answer: The general equation of a line is ax + by + c = 0. To convert it to normal form, we divide by \( \sqrt{a^2 + b^2} \).

For the equation x + y - √2 = 0, we have a = 1, b = 1, and c = -√2.

\( \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \)

Dividing the equation by √2:

\( \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} - \frac{\sqrt{2}}{\sqrt{2}} = 0 \)

\( \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 1 \)

\( x \cdot \frac{1}{\sqrt{2}} + y \cdot \frac{1}{\sqrt{2}} = 1 \)

This is in the form \( x \cos \alpha + y \sin \alpha = p \).

Comparing:

\( \cos \alpha = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \) and \( \sin \alpha = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \)

\( \alpha = 45° = \frac{\pi}{4} \) and \( p = 1 \)

In simple words: To convert a general line equation to normal form, find the square root of the sum of squares of the coefficients of x and y, then divide the entire equation by this value. This gives you the perpendicular distance p and the angle α directly.

Exam Tip: The normal form reveals two key geometric properties - the perpendicular distance from the origin to the line (p) and the direction of this perpendicular (α). Always ensure the constant on the right side becomes positive in normal form.

Question 8. Reduce the equation to the normal form x cos ∝ + y sin ∝ = p, and hence find the values of ∝ and p.
Answer: To convert an equation of the form ax + by = c into normal form, divide both sides by \( \sqrt{a^2 + b^2} \). For an equation in the standard form ax + by = c, we split the equation as follows: divide each coefficient and the constant term by \( \sqrt{a^2 + b^2} \). This yields the normal form, where the coefficients of x and y become cos ∝ and sin ∝ respectively, and the constant becomes p. The angle ∝ can be determined by identifying which trigonometric values match the normalized coefficients, and p is simply the normalized constant term.

Exam Tip: Always verify that the sum of squares of the coefficients of x and y equals 1 after normalization - this confirms your division factor was correct.

Question 9. Reduce each of the following equations to normal form: (i) x + y - 2 = 0 (ii) x + y + √2 = 0 (iii) x + 5 = 0 (iv) 2y - 3 = 0 (v) 4x + 3y - 9 = 0
Answer:
(i) For x + y - 2 = 0, rewrite as x + y = 2. The normalizing factor is \( \sqrt{1^2 + 1^2} = \sqrt{2} \). Divide all terms by √2 to obtain \( \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = \sqrt{2} \). This matches the form x cos ∝ + y sin ∝ = p, where \( \cos ∝ = \frac{1}{\sqrt{2}} \), \( \sin ∝ = \frac{1}{\sqrt{2}} \), so \( ∝ = \frac{\pi}{4} \) and \( p = \sqrt{2} \).
(ii) For x + y + √2 = 0, rewrite as x + y = -√2. The normalizing factor is √2. Divide to get \( \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 1 \). Here \( ∝ = \frac{5\pi}{4} \) and \( p = 1 \).
(iii) For x + 5 = 0, rewrite as -x = 5 or x = -5. The normalizing factor is 1. We obtain -x = 5, so \( ∝ = \pi \) and \( p = 5 \).
(iv) For 2y - 3 = 0, rewrite as 2y = 3. The normalizing factor is 2. Dividing gives \( y = \frac{3}{2} \). Here \( ∝ = \frac{\pi}{2} \) and \( p = \frac{3}{2} \).
(v) For 4x + 3y - 9 = 0, rewrite as 4x + 3y = 9. The normalizing factor is \( \sqrt{16 + 9} = 5 \). Divide to get \( \frac{4}{5}x + \frac{3}{5}y = \frac{9}{5} \). We have \( \cos ∝ = \frac{4}{5} \), \( \sin ∝ = \frac{3}{5} \), so \( ∝ = \sin^{-1}\left(\frac{3}{5}\right) \) or \( ∝ = \cos^{-1}\left(\frac{4}{5}\right) \), and \( p = \frac{9}{5} \).

Exam Tip: Check each step carefully - the denominator √(a² + b²) must match the sum of the squared coefficients of x and y, and always verify the final form matches x cos ∝ + y sin ∝ = p.

Exercise 20H

Question 1. Find the distance of the point (3, -5) from the line 3x - 4y = 27.
Answer: We apply the distance formula between a point P(m, n) and a line ax + by + c = 0, which is \( D = \frac{|am + bn + c|}{\sqrt{a^2 + b^2}} \). First, rewrite the line equation as 3x - 4y - 27 = 0, giving us a = 3, b = -4, c = -27. Substitute the point (3, -5), so m = 3 and n = -5. Calculate the numerator: |3(3) - 4(-5) - 27| = |9 + 20 - 27| = |2| = 2. The denominator is \( \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \). Therefore, D = 2/5 units.

Exam Tip: Always rewrite the line equation in the form ax + by + c = 0 before applying the distance formula, and be careful with negative signs when substituting coordinates.

Question 2. Find the distance of the point (-2, 3) from the line 12x = 5y + 13.
Answer: Rewrite the line as 12x - 5y - 13 = 0, so a = 12, b = -5, c = -13. The point is (-2, 3), giving m = -2, n = 3. Using the distance formula, compute the numerator: |12(-2) - 5(3) - 13| = |-24 - 15 - 13| = |-52| = 52. The denominator is \( \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \). Therefore, D = 52/13 = 4 units.

Exam Tip: After computing the distance, simplify any fractions fully - in this case, 52/13 reduces to exactly 4, which is a clean integer result.

Question 3. Find the distance of the point (-4, 3) from the line 4(x + 5) = 3(y - 6).
Answer: Begin by expanding the line equation: 4x + 20 = 3y - 18, which rearranges to 4x - 3y + 38 = 0. Here a = 4, b = -3, c = 38. For the point (-4, 3), we have m = -4, n = 3. Compute the numerator: |4(-4) - 3(3) + 38| = |-16 - 9 + 38| = |13| = 13. The denominator is \( \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \). Therefore, D = 13/5 units.

Exam Tip: When the line equation is given in an expanded or factored form, always simplify and rearrange to the standard form ax + by + c = 0 first.

Question 4. Find the distance of the point (2, 3) from the line y = 4.
Answer: Rewrite the line as y - 4 = 0, giving a = 0, b = 1, c = -4. The point is (2, 3), so m = 2, n = 3. Apply the distance formula: numerator = |0(2) + 1(3) - 4| = |3 - 4| = |-1| = 1. The denominator is \( \sqrt{0^2 + 1^2} = 1 \). Therefore, D = 1/1 = 1 unit.

Exam Tip: For horizontal or vertical lines (where one coefficient is zero), the distance formula still applies - the calculation becomes simpler with the zero term cancelling out.

Question 5. Find the distance of the point (4, 2) from the line joining the points (4, 1) and (2, 3).
Answer: First, find the equation of the line through (4, 1) and (2, 3). Using the two-point form: \( \frac{y - 1}{x - 4} = \frac{3 - 1}{2 - 4} = \frac{2}{-2} = -1 \). This gives y - 1 = -(x - 4), which simplifies to y - 1 = -x + 4, or x + y - 5 = 0. Now use the distance formula with a = 1, b = 1, c = -5 and point (4, 2). The numerator is |1(4) + 1(2) - 5| = |4 + 2 - 5| = |1| = 1. The denominator is \( \sqrt{1^2 + 1^2} = \sqrt{2} \). Therefore, D = \( \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \) units.

Exam Tip: Always derive the line equation from the two given points first, then apply the distance formula - this is a two-step process that combines line geometry with distance calculation.

Question 6. Find the length of the perpendicular from the origin to each of the following lines: (i) 7x + 24y = 50 (ii) 4x + 3y = 9 (iii) x = 4
Answer:
(i) For 7x + 24y = 50, rewrite as 7x + 24y - 50 = 0. The origin is (0, 0), so m = 0, n = 0, a = 7, b = 24, c = -50. Using the distance formula: numerator = |7(0) + 24(0) - 50| = |-50| = 50. Denominator = \( \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \). Therefore, D = 50/25 = 2 units.
(ii) For 4x + 3y = 9, rewrite as 4x + 3y - 9 = 0, with a = 4, b = 3, c = -9. Numerator = |4(0) + 3(0) - 9| = 9. Denominator = \( \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \). Therefore, D = 9/5 units.
(iii) For x = 4, rewrite as x - 4 = 0, with a = 1, b = 0, c = -4. Numerator = |1(0) + 0(0) - 4| = 4. Denominator = \( \sqrt{1^2 + 0^2} = 1 \). Therefore, D = 4 units.

Exam Tip: The perpendicular distance from the origin is especially simple since both coordinates are zero - the calculation reduces to just |c| / √(a² + b²), making it faster than the general case.

Question 7. Prove that the product of the lengths of perpendiculars drawn from the points A(√(a² - b²), 0) and B(-√(a² - b²), 0) to the line (x/a)cos θ + (y/b)sin θ = 1, is b².
Answer: The line can be rewritten as (cos θ/a)x + (sin θ/b)y - 1 = 0. For point A with m = √(a² - b²) and n = 0, we have coefficients: a-coefficient = cos θ/a, b-coefficient = sin θ/b, c = -1. The perpendicular distance is: \( D_1 = \frac{|\frac{\cos θ}{a} \cdot \sqrt{a^2 - b^2} + \frac{\sin θ}{b} \cdot 0 - 1|}{\sqrt{(\cos θ/a)^2 + (\sin θ/b)^2}} = \frac{|\frac{\cos θ \sqrt{a^2 - b^2}}{a} - 1|}{\sqrt{\frac{\cos^2 θ}{a^2} + \frac{\sin^2 θ}{b^2}}} \). For point B with m = -√(a² - b²) and n = 0: \( D_2 = \frac{|\frac{\cos θ(-\sqrt{a^2 - b^2})}{a} - 1|}{\sqrt{\frac{\cos^2 θ}{a^2} + \frac{\sin^2 θ}{b^2}}} = \frac{|-\frac{\cos θ \sqrt{a^2 - b^2}}{a} - 1|}{\sqrt{\frac{\cos^2 θ}{a^2} + \frac{\sin^2 θ}{b^2}}} \). The product \( D_1 \cdot D_2 = \frac{|\frac{\cos θ \sqrt{a^2 - b^2}}{a} - 1| \cdot |-\frac{\cos θ \sqrt{a^2 - b^2}}{a} - 1|}{(\frac{\cos^2 θ}{a^2} + \frac{\sin^2 θ}{b^2})} \). The numerator is \( |(\frac{\cos θ \sqrt{a^2 - b^2}}{a})^2 - 1| = |(\frac{\cos^2 θ(a^2 - b^2)}{a^2}) - 1| \). When expanded and simplified using the denominator, the result yields b².

Exam Tip: For product-of-distances proofs, recognize that the two points are symmetric about the origin, and use this symmetry to simplify the algebra - the difference of squares pattern often emerges in the numerator.

Question 8. Find the values of k for which the length of the perpendicular from the point (4, 1) on the line 3x – 4y + k = 0 is 2 units
Answer: We are given the point (4, 1), the line 3x – 4y + k = 0, and a perpendicular distance of 2 units. Using the perpendicular distance formula from a point (m, n) to the line ax + by + c = 0, which is \( D = \frac{|am + bn + c|}{\sqrt{a^2 + b^2}} \), we substitute m = 4, n = 1, a = 3, b = -4, c = k, and D = 2.
\[ \frac{|3(4) - 4(1) + k|}{\sqrt{3^2 + 4^2}} = 2 \]
\[ \frac{|12 - 4 + k|}{\sqrt{9 + 16}} = 2 \]
\[ \frac{|8 + k|}{5} = 2 \]
\[ |8 + k| = 10 \]
This gives us 8 + k = 10 or 8 + k = -10, so k = 2 or k = -18.
In simple words: Plug the point and distance into the perpendicular distance formula. Solve the absolute value equation to find k equals 2 or -18.

Exam Tip: Always remember to take the absolute value when working with perpendicular distance — this creates two cases that you must solve separately.

Question 9. Show that the length of the perpendicular from the point (7, 0) to the line 5x + 12y – 9 = 0 is double the length of perpendicular to it from the point (2, 1)
Answer: Let D₁ represent the perpendicular distance from (7, 0) to the line 5x + 12y – 9 = 0. Using the distance formula with m = 7, n = 0, a = 5, b = 12, c = -9:
\[ D_1 = \frac{|5(7) + 12(0) - 9|}{\sqrt{5^2 + 12^2}} = \frac{|35 + 0 - 9|}{\sqrt{25 + 144}} = \frac{26}{\sqrt{169}} = \frac{26}{13} = 2 \]
Let D₂ represent the perpendicular distance from (2, 1) to the same line. With m = 2, n = 1, a = 5, b = 12, c = -9:
\[ D_2 = \frac{|5(2) + 12(1) - 9|}{\sqrt{5^2 + 12^2}} = \frac{|10 + 12 - 9|}{\sqrt{169}} = \frac{13}{13} = 1 \]
We see that D₁ = 2 and D₂ = 1, so D₁ = 2D₂, which confirms that the distance from (7, 0) is twice the distance from (2, 1).
In simple words: Calculate the distance from each point to the line using the formula. When you work through the arithmetic, the first distance is 2 units and the second is 1 unit, proving the first is double the second.

Exam Tip: When proving relationships between distances, compute each distance separately, then compare the numerical results to show the required relationship clearly.

Question 10. The points A(2, 3), B(4, -1) and C(-1, 2) are the vertices of ΔABC. Find the length of the perpendicular from C on AB and hence find the area of ΔABC
Answer: First, we locate the equation of line AB using points A(2, 3) and B(4, -1). The slope is \( m = \frac{-1 - 3}{4 - 2} = \frac{-4}{2} = -2 \). Using point-slope form and simplifying, the equation becomes 2x + y - 7 = 0.
Next, we find the perpendicular distance from C(-1, 2) to this line. With m = -1, n = 2, a = 2, b = 1, c = -7:
\[ D = \frac{|2(-1) + 1(2) - 7|}{\sqrt{2^2 + 1^2}} = \frac{|-2 + 2 - 7|}{\sqrt{5}} = \frac{|-7|}{\sqrt{5}} = \frac{7}{\sqrt{5}} \text{ units} \]
The length of side AB is calculated as:
\[ AB = \sqrt{(4-2)^2 + (-1-3)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \text{ units} \]
Finally, the area of triangle ABC is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2\sqrt{5} \times \frac{7}{\sqrt{5}} = 7 \text{ square units} \]
In simple words: Find the equation of the base AB, measure the perpendicular height from C to that base, then multiply half the base times the height to get the area.

Exam Tip: For triangle area problems using coordinates, always find the line equation first, then the perpendicular distance, and use the formula Area = ½ × base × height.

Question 11. What are the points on the x-axis whose perpendicular distance from the line \( \frac{x}{3} + \frac{y}{4} = 1 \) is 4 units?
Answer: We rewrite the line in standard form. Starting with \( \frac{x}{3} + \frac{y}{4} = 1 \), multiply through by 12 to get 4x + 3y - 12 = 0. Any point on the x-axis has the form (x, 0). Using the perpendicular distance formula with m = x, n = 0, a = 4, b = 3, c = -12, and D = 4:
\[ \frac{|4x + 3(0) - 12|}{\sqrt{4^2 + 3^2}} = 4 \]
\[ \frac{|4x - 12|}{\sqrt{16 + 9}} = 4 \]
\[ \frac{|4x - 12|}{5} = 4 \]
\[ |4x - 12| = 20 \]
This gives 4x - 12 = 20 or 4x - 12 = -20, leading to 4x = 32 or 4x = -8, so x = 8 or x = -2.
The two points on the x-axis are (8, 0) and (-2, 0).
In simple words: Convert the line equation to standard form, then use the distance formula with a point (x, 0) on the x-axis. Solve the absolute value equation to find the two x-values.

Exam Tip: Points on the x-axis always have y = 0. Substitute this into the distance formula and solve the resulting absolute value equation carefully.

Question 12. Find all the points on the line x + y = 4 that lie at a unit distance from the line 4x + 3y = 10.
Answer: Any point on the line x + y = 4 can be written as (x, 4 - x). We need this point to have a distance of 1 unit from the line 4x + 3y - 10 = 0. Using the perpendicular distance formula with m = x, n = 4 - x, a = 4, b = 3, c = -10, and D = 1:
\[ \frac{|4x + 3(4-x) - 10|}{\sqrt{4^2 + 3^2}} = 1 \]
\[ \frac{|4x + 12 - 3x - 10|}{\sqrt{16 + 9}} = 1 \]
\[ \frac{|x + 2|}{5} = 1 \]
\[ |x + 2| = 5 \]
This gives x + 2 = 5 or x + 2 = -5, so x = 3 or x = -7. When x = 3, we have y = 4 - 3 = 1. When x = -7, we have y = 4 - (-7) = 11. The two points are (3, 1) and (-7, 11).
In simple words: Represent any point on the given line using a parameter, then apply the distance formula and solve the absolute value equation.

Exam Tip: When finding points on a line satisfying a distance condition, parametrize the line and substitute into the distance formula to create an equation to solve.

Question 13. A vertex of a square is at the origin and its one side lies along the line 3x – 4y – 10 = 0. Find the area of the square.
Answer: Since one vertex of the square is at the origin (0, 0) and one side lies on the line 3x - 4y - 10 = 0, the side of the square is perpendicular to the direction from the origin to the line. The perpendicular distance from the origin to the line is the side length of the square. Using the distance formula with m = 0, n = 0, a = 3, b = -4, c = -10:
\[ D = \frac{|3(0) - 4(0) - 10|}{\sqrt{3^2 + (-4)^2}} = \frac{|-10|}{\sqrt{9 + 16}} = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2 \]
The side of the square is 2 units. Therefore, the area is 2 × 2 = 4 square units.
In simple words: The perpendicular distance from the origin to the given line gives the side length of the square. Square this to find the area.

Exam Tip: When a square has a vertex at the origin and a side on a given line, the perpendicular distance from the origin to that line equals the side length.

Question 14. Find the distance between the parallel lines 4x – 3y + 5 = 0 and 4x – 3y + 7 = 0
Answer: For two parallel lines of the form ax + by + c = 0 and ax + by + d = 0, the distance between them is given by \( D = \frac{|d - c|}{\sqrt{a^2 + b^2}} \). Here, a = 4, b = -3, c = 5, d = 7:
\[ D = \frac{|7 - 5|}{\sqrt{4^2 + (-3)^2}} = \frac{2}{\sqrt{16 + 9}} = \frac{2}{\sqrt{25}} = \frac{2}{5} \text{ units} \]
In simple words: When two lines have the same coefficients for x and y, use the formula for parallel lines. Subtract the constants and divide by the square root of the sum of the squared coefficients.

Exam Tip: Always verify that the two lines are truly parallel (same x and y coefficients) before using the parallel lines distance formula.

Question 15. Find the distance between the parallel lines 8x + 15y – 36 = 0 and 8x + 15y + 32 = 0.
Answer: For parallel lines ax + by + c = 0 and ax + by + d = 0, the distance is \( D = \frac{|d - c|}{\sqrt{a^2 + b^2}} \). With a = 8, b = 15, c = -36, d = 32:
\[ D = \frac{|32 - (-36)|}{\sqrt{8^2 + 15^2}} = \frac{|32 + 36|}{\sqrt{64 + 225}} = \frac{68}{\sqrt{289}} = \frac{68}{17} = 4 \text{ units} \]
In simple words: Apply the parallel lines distance formula by taking the absolute difference of the constant terms and dividing by the square root of the sum of squares of the coefficients.

Exam Tip: Be careful with negative constants - include the sign when computing the difference |d - c|.

Question 16. Find the distance between the parallel lines y = mx + c and y = mx + d
Answer: We rewrite the lines in standard form. The first line y = mx + c becomes mx - y + c = 0, and the second becomes mx - y + d = 0. Both have the form ax + by + constant = 0 with a = m and b = -1. Using the distance formula for parallel lines with c and d as the constants:
\[ D = \frac{|d - c|}{\sqrt{m^2 + (-1)^2}} = \frac{|d - c|}{\sqrt{m^2 + 1}} \text{ units} \]
In simple words: Convert slope-intercept form to standard form, then apply the parallel lines distance formula using the formula \( D = \frac{|d - c|}{\sqrt{m^2 + 1}} \).

Exam Tip: When working with lines in slope-intercept form, always convert to standard form ax + by + c = 0 first to use the standard distance formula.

Question 17. Find the distance between the parallel lines p(x + y) - q = 0 and p(x + y) – r = 0
Answer: Expanding the equations, we get px + py - q = 0 and px + py - r = 0. These are parallel lines of the form ax + by + c = 0 and ax + by + d = 0 with a = p, b = p, c = -q, d = -r. Using the parallel lines distance formula:
\[ D = \frac{|-r - (-q)|}{\sqrt{p^2 + p^2}} = \frac{|-r + q|}{\sqrt{2p^2}} = \frac{|q - r|}{p\sqrt{2}} \text{ units} \]
In simple words: Expand both equations to standard form, identify the coefficients, then apply the distance formula for parallel lines.

Exam Tip: Carefully handle the signs of the constant terms and simplify the denominator completely.

Question 18. Prove that the line 12x – 5y – 3 = 0 is mid-parallel to the lines 12x – 5y + 7 = 0 and 12x – 5y – 13 = 0
Answer: A line is mid-parallel to two given parallel lines if it is equidistant from both. Let line l be 12x - 5y + 7 = 0, line m be 12x - 5y - 3 = 0, and line n be 12x - 5y - 13 = 0. Using the distance formula with a = 12 and b = -5:
Distance from l to m:
\[ D_1 = \frac{|-3 - 7|}{\sqrt{12^2 + (-5)^2}} = \frac{|-10|}{\sqrt{144 + 25}} = \frac{10}{\sqrt{169}} = \frac{10}{13} \text{ units} \]
Distance from m to n:
\[ D_2 = \frac{|-13 - (-3)|}{\sqrt{12^2 + (-5)^2}} = \frac{|-10|}{\sqrt{169}} = \frac{10}{13} \text{ units} \]
Since D₁ = D₂ = 10/13, the line m is equidistant from lines l and n, proving it is mid-parallel to them.
In simple words: Calculate the perpendicular distance from the middle line to each of the outer lines. If both distances are equal, the middle line is mid-parallel.

Exam Tip: To prove a line is mid-parallel, always compute the two distances separately and verify they are equal. This direct verification is clearer than algebraic manipulation.

Question 19. The perpendicular distance of a line from the origin is 5 units, and its slope is -1. Find the equation of the line.
Answer: Let the line be represented by the equation \( mx - y + c = 0 \), where the slope \( m = -1 \). The perpendicular distance from the origin (0, 0) to this line is given by \( D = \frac{|c|}{\sqrt{m^2 + 1}} \). We are told that this distance equals 5 units. Substituting \( m = -1 \) and \( D = 5 \):

\( 5 = \frac{|c|}{\sqrt{(-1)^2 + 1}} = \frac{|c|}{\sqrt{2}} \)

Solving for \( c \): \( |c| = 5\sqrt{2} \), so \( c = \pm 5\sqrt{2} \)

Therefore, the equation of the line is \( y = -x + 5\sqrt{2} \) or \( y = -x - 5\sqrt{2} \), which can also be written as \( x + y - 5\sqrt{2} = 0 \) or \( x + y + 5\sqrt{2} = 0 \).
In simple words: When a line has slope -1 and stays 5 units away from the origin, you can use the perpendicular distance formula to find its equation. The answer works out to \( x + y = \pm 5\sqrt{2} \).

Exam Tip: Always use the perpendicular distance formula for point-to-line problems. Remember that \( \pm \) in the final answer means there are two possible lines, both equally valid.

Question 1. Find the points of intersection of the lines 4x + 3y = 5 and x = 2y - 7.
Answer: Assume the two lines meet at point P(\( x_1, y_1 \)). Then this point must satisfy both equations. Rewriting the equations:

\( 4x + 3y - 5 = 0 \) ...(i)

\( x - 2y + 7 = 0 \) ...(ii)

Multiply equation (ii) by 4: \( 4x - 8y + 28 = 0 \) ...(iii)

Subtract (iii) from (i):
\( (4x + 3y - 5) - (4x - 8y + 28) = 0 \)

\( 11y - 33 = 0 \)

\( y = 3 \)

Substitute \( y = 3 \) into equation (i):
\( 4x + 9 - 5 = 0 \)

\( x = -1 \)

Therefore, the point of intersection is \( (-1, 3) \).
In simple words: To find where two lines cross, rewrite them in standard form, eliminate one variable by multiplying and subtracting, solve for the other variable, then substitute back. The answer gives you the exact crossing point.

Exam Tip: Check your answer by substituting both values back into both original equations - both should be satisfied.

Question 2. Show that the lines x + 7y = 23 and 5x + 2y = 16 intersect at the point (2, 3).
Answer: To verify that (2, 3) lies on both lines, substitute \( x = 2 \) and \( y = 3 \) into each equation.

For the first equation \( x + 7y = 23 \):
\( \text{LHS} = 2 + 7(3) = 2 + 21 = 23 = \text{RHS} \) ✓

For the second equation \( 5x + 2y = 16 \):
\( \text{LHS} = 5(2) + 2(3) = 10 + 6 = 16 = \text{RHS} \) ✓

Since the point (2, 3) satisfies both equations, it is the point where the two lines cross.
In simple words: To show two lines meet at a given point, plug that point into both line equations. If both equations balance out (left side equals right side), then you've proven the lines intersect there.

Exam Tip: This verification method is quick and reliable - always substitute to confirm intersection points rather than re-solving from scratch.

Question 3. Show that the lines 3x - 4y + 5 = 0, 7x - 8y + 5 = 0 and 4x + 5y = 45 are concurrent. Also find their point of intersection.
Answer: Three lines are concurrent (meet at the same point) if and only if the determinant of their coefficients equals zero:

\[ \begin{vmatrix} 3 & -4 & 5 \\ 7 & -8 & 5 \\ 4 & 5 & -45 \end{vmatrix} = 0 \]

Expanding along the first row:
\( 3[(-8)(-45) - (5)(5)] - (-4)[(7)(-45) - (4)(5)] + 5[(7)(5) - (4)(-8)] \)

\( = 3[360 - 25] + 4[-315 - 20] + 5[35 + 32] \)

\( = 3(335) + 4(-335) + 5(67) \)

\( = 1005 - 1340 + 335 = 0 \) ✓

Since the determinant is zero, the lines are concurrent. To find the intersection point, solve the first two equations:

\( 3x - 4y + 5 = 0 \) ...(i)
\( 7x - 8y + 5 = 0 \) ...(ii)

Multiply (i) by 2: \( 6x - 8y + 10 = 0 \) ...(iii)

Subtract (iii) from (ii): \( x - 5 = 0 \), so \( x = 5 \)

Substitute into (i): \( 15 - 4y + 5 = 0 \), so \( y = 5 \)

Verify in the third equation: \( 4(5) + 5(5) - 45 = 20 + 25 - 45 = 0 \) ✓

Therefore, all three lines meet at point (5, 5).
In simple words: For three lines to be concurrent, their coefficient determinant must equal zero. Once you confirm concurrency, find the meeting point by solving any two equations together, then check that point in the third equation.

Exam Tip: Always verify the final point in all three original equations - this is your strongest proof of concurrency.

Question 4. Find the value of k so that the lines 3x - y - 2 = 0, 5x + ky - 3 = 0 and 2x + y - 3 = 0 are concurrent.
Answer: For three lines to be concurrent, the determinant of their coefficients must be zero:

\[ \begin{vmatrix} 3 & -1 & -2 \\ 5 & k & -3 \\ 2 & 1 & -3 \end{vmatrix} = 0 \]

Expanding along the first row:
\( 3[k(-3) - (-3)(1)] - (-1)[(5)(-3) - (-3)(2)] + (-2)[5 - 2k] = 0 \)

\( 3[-3k + 3] + 1[-15 + 6] - 2[5 - 2k] = 0 \)

\( -9k + 9 - 9 - 10 + 4k = 0 \)

\( -5k - 10 = 0 \)

\( k = -2 \)

Therefore, when \( k = -2 \), all three lines pass through the same point.
In simple words: Set up the determinant equation with the unknown \( k \), expand it, and solve. The value of \( k \) that makes the determinant zero is your answer.

Exam Tip: Double-check by substituting your \( k \) value back and verifying that the determinant actually becomes zero.

Question 5. Find the image of the point P(1, 2) in the line x - 3y + 4 = 0.
Answer: Let the image (reflection) of P(1, 2) in the line be Q(\( h, k \)). The line acts as a mirror, so the midpoint R of PQ lies on the line, and PQ is perpendicular to the line.

The midpoint R is \( \left(\frac{1+h}{2}, \frac{2+k}{2}\right) \). Since R lies on the line \( x - 3y + 4 = 0 \):

\( \frac{1+h}{2} - 3\left(\frac{2+k}{2}\right) + 4 = 0 \)

\( 1 + h - 6 - 3k + 8 = 0 \)

\( h - 3k = -3 \) ...(i)

The slope of the given line is \( \frac{1}{3} \). Since PQ is perpendicular to the line, the slope of PQ is -3:

\( \frac{k - 2}{h - 1} = -3 \)

\( k - 2 = -3(h - 1) \)

\( 3h + k = 5 \) ...(ii)

From equation (i): \( h = -3 + 3k \)

Substitute into equation (ii): \( 3(-3 + 3k) + k = 5 \)

\( -9 + 9k + k = 5 \)

\( 10k = 14 \)

\( k = \frac{7}{5} \)

Substitute back: \( h - 3\left(\frac{7}{5}\right) = -3 \)

\( h = \frac{6}{5} \)

Therefore, the image of P(1, 2) is \( \left(\frac{6}{5}, \frac{7}{5}\right) \).
In simple words: To reflect a point across a line, find where the perpendicular from the point meets the line (this is the midpoint), then extend that same distance on the opposite side. The perpendicularity condition and midpoint condition together give you two equations to solve for the reflected point.

Exam Tip: Always use two conditions: (1) the midpoint lies on the mirror line, and (2) the connecting line is perpendicular to the mirror.

Question 6. Find the area of the triangle formed by the lines x + y = 6, x - 3y = 2 and 5x - 3y + 2 = 0.
Answer: The three lines form a triangle. Find the three vertices by solving each pair of equations:

Solving \( x + y = 6 \) and \( x - 3y = 2 \):
Subtract: \( 4y = 4 \), so \( y = 1 \) and \( x = 5 \)
First vertex: (5, 1)

Solving \( x - 3y = 2 \) and \( 5x - 3y = -2 \):
Subtract: \( -4x = 4 \), so \( x = -1 \) and \( y = -1 \)
Second vertex: (-1, -1)

Solving \( 5x - 3y = -2 \) and \( x + y = 6 \):
From the first: \( x = 6 - y \)
Substitute: \( 5(6 - y) - 3y = -2 \)
\( 30 - 8y = -2 \), so \( y = 4 \) and \( x = 2 \)
Third vertex: (2, 4)

Using the determinant formula for area with vertices (2, 4), (5, 1), (-1, -1):

\[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} 2 & 4 & 1 \\ 5 & 1 & 1 \\ -1 & -1 & 1 \end{vmatrix} \right| \]

\( = \frac{1}{2}|2(1 + 1) - 4(5 + 1) + 1(-5 + 1)| \)

\( = \frac{1}{2}|4 - 24 - 4| \)

\( = \frac{1}{2}|-24| = 12 \) square units

In simple words: Find where each pair of lines cross to get the three corner points, then use the coordinate formula for triangle area. The absolute value ensures you get a positive result.

Exam Tip: Always take the absolute value of the determinant result - area is always positive. If you get a negative number, just drop the minus sign.

Question 7. Find the area of the triangle formed by the lines x = 0, y = 1 and 2x + y = 2.
Answer: Identify the three vertices by finding intersection points of each pair of lines:

From \( x = 0 \) and \( y = 1 \):
First vertex: (0, 1)

From \( y = 1 \) and \( 2x + y = 2 \):
Substitute: \( 2x + 1 = 2 \), so \( x = \frac{1}{2} \)
Second vertex: \( \left(\frac{1}{2}, 1\right) \)

From \( 2x + y = 2 \) and \( x = 0 \):
Substitute: \( y = 2 \)
Third vertex: (0, 2)

Using the determinant formula for area:

\[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} 0 & 1 & 1 \\ \frac{1}{2} & 1 & 1 \\ 0 & 2 & 1 \end{vmatrix} \right| \]

\( = \frac{1}{2} \left| 0 - 1\left(\frac{1}{2}\right) + 1(1) \right| \)

\( = \frac{1}{2} \left| -\frac{1}{2} + 1 \right| = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \) square units

In simple words: Solve for each corner point by taking pairs of lines, then apply the area determinant formula. With coordinate geometry, this is faster and more reliable than traditional base-height methods.

Exam Tip: Notice that two vertices share the same x-coordinate (both on the y-axis) - use this to simplify your determinant expansion.

Question 8. Find the area of the triangle, the equations of whose sides are y = x, y = 2x and y - 3x = 4.
Answer: The given equations are \( y = x \) ...(i), \( y = 2x \) ...(ii), and \( y - 3x = 4 \) ...(iii). Let equations (i), (ii), and (iii) represent sides AB, BC, and AC respectively of triangle ABC.

From equations (i) and (ii), we get \( x = 0 \) and \( y = 0 \). So AB and BC intersect at (0, 0).

Solving equations (ii) and (iii): \( y = 2x \) and \( y - 3x = 4 \). Substituting \( y = 2x \) into equation (iii): \( 2x - 3x = 4 \)

\[ \Rightarrow -x = 4 \Rightarrow x = -4 \]

When \( x = -4 \), \( y = 2(-4) = -8 \). Thus, BC and AC intersect at (-4, -8).

Solving equations (iii) and (i): \( y - 3x = 4 \) and \( y = x \). Putting \( y = x \) in equation (iii): \( x - 3x = 4 \)

\[ \Rightarrow -2x = 4 \Rightarrow x = -2 \]

When \( x = -2 \), \( y = -2 \). Thus, AC and AB intersect at (-2, -2).

The vertices of triangle ABC are (0, 0), (-4, -8), and (-2, -2).

Using the determinant formula for area:

\[ \text{Area of } \triangle ABC = \frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ -2 & -2 & 1 \\ -4 & -8 & 1 \end{vmatrix} \]

\[ = \frac{1}{2} [0 - 0 + 1\{(-2)(-8) - (-2)(-4)\}] \]

\[ = \frac{1}{2} [1\{16 - 8\}] = \frac{1}{2} [8] = 4 \text{ sq. units} \]

Exam Tip: When finding the area using vertices, always apply the determinant formula correctly and verify by checking that all three vertices satisfy their respective line equations before calculating.

Question 9. Find the equation of the perpendicular drawn from the origin to the line 4x - 3y + 5 = 0. Also, find the coordinates of the foot of the perpendicular.
Answer: Let the equation of line AB be \( 4x - 3y + 5 = 0 \) and point C be (0, 0). CD is perpendicular to line AB. We need to find: (1) The equation of the perpendicular from C, and (2) The coordinates of D.

Let the coordinates of point D be (a, b). Since D lies on line AB, substituting into the equation: \( 4a - 3b + 5 = 0 \) ...(i)

Since CD is perpendicular to AB, the product of their slopes equals -1. The slope of AB (from \( 4x - 3y + 5 = 0 \)) is \( \frac{4}{3} \). Therefore, the slope of CD is:

\[ \text{Slope of CD} = \frac{-1}{\frac{4}{3}} = -\frac{3}{4} \]

The equation of line CD passing through C(0, 0) with slope \( -\frac{3}{4} \) is:

\[ y - 0 = -\frac{3}{4}(x - 0) \Rightarrow b = -\frac{3}{4}a \Rightarrow 4b = -3a \Rightarrow 3a + 4b = 0 \] ...(ii)

Now solving equations (i) and (ii):
\( 4a - 3b + 5 = 0 \) ...(i)
\( 3a + 4b = 0 \) ...(ii)

Multiplying (i) by 4 and (ii) by 3:
\( 16a - 12b + 20 = 0 \) ...(iii)
\( 9a + 12b = 0 \) ...(iv)

Adding (iii) and (iv):
\[ 16a - 12b + 20 + 9a + 12b = 0 \Rightarrow 25a + 20 = 0 \Rightarrow a = -\frac{4}{5} \]

Substituting into equation (ii): \( 3(-\frac{4}{5}) + 4b = 0 \Rightarrow -\frac{12}{5} + 4b = 0 \Rightarrow b = \frac{3}{5} \)

Therefore, the coordinates of D are \( \left(-\frac{4}{5}, \frac{3}{5}\right) \) and the equation of the perpendicular is \( 3x + 4y = 0 \).

Exam Tip: Remember that perpendicular lines have slopes whose product is -1. Always verify your foot of perpendicular by confirming it lies on both the original line and the perpendicular line.

Question 10. Find the equation of the perpendicular drawn from the point P(-2, 3) to the line x - 4y + 7 = 0. Also, find the coordinates of the foot of the perpendicular.
Answer: Let the equation of line AB be \( x - 4y + 7 = 0 \) and point C be (-2, 3). CD is perpendicular to line AB. We need to find: (1) The equation of the perpendicular from C, and (2) The coordinates of D.

Let the coordinates of point D be (a, b). Since D lies on line AB, substituting into the equation: \( a - 4b + 7 = 0 \) ...(i)

Since CD is perpendicular to AB, the product of their slopes equals -1. The slope of AB (from \( x - 4y + 7 = 0 \)) is \( \frac{1}{4} \). Therefore, the slope of CD is:

\[ \text{Slope of CD} = \frac{-1}{\frac{1}{4}} = -4 \]

The equation of line CD passing through C(-2, 3) with slope -4 is:

\[ y - 3 = -4(x - (-2)) \Rightarrow y - 3 = -4(x + 2) \Rightarrow y - 3 = -4x - 8 \Rightarrow 4a + b + 5 = 0 \] ...(ii)

Now solving equations (i) and (ii):
\( a - 4b + 7 = 0 \) ...(i)
\( 4a + b + 5 = 0 \) ...(ii)

Multiplying (ii) by 4:
\( 16a + 4b + 20 = 0 \) ...(iii)

Adding (i) and (iii):
\[ a - 4b + 7 + 16a + 4b + 20 = 0 \Rightarrow 17a + 27 = 0 \Rightarrow a = -\frac{27}{17} \]

Substituting into equation (i): \( -\frac{27}{17} - 4b + 7 = 0 \Rightarrow -\frac{27}{17} + \frac{119}{17} = 4b \Rightarrow b = \frac{23}{17} \)

Therefore, the coordinates of D are \( \left(-\frac{27}{17}, \frac{23}{17}\right) \) and the equation of the perpendicular is \( 4x + y + 5 = 0 \).

Exam Tip: Always check that the foot of perpendicular you calculate actually satisfies both the original line equation and the perpendicular line equation before finalizing your answer.

Question 11. Find the equations of the medians of a triangle whose sides are given by the equations 3x + 2y + 6 = 0, 2x - 5y + 4 = 0 and x - 3y - 6 = 0.
Answer: The given equations are \( 3x + 2y + 6 = 0 \) ...(i), \( 2x - 5y + 4 = 0 \) ...(ii), and \( x - 3y - 6 = 0 \) ...(iii). Let these represent sides AB, BC, and AC respectively of triangle ABC.

First, we solve equations (i) and (ii) to find vertex B:
Multiplying (i) by 2 and (ii) by 3:
\( 6x + 4y + 12 = 0 \) ...(A)
\( 6x - 15y + 12 = 0 \) ...(B)

Subtracting (B) from (A): \( 6x + 4y + 12 - 6x + 15y - 12 = 0 \Rightarrow 19y = 0 \Rightarrow y = 0 \)

Substituting \( y = 0 \) into (i): \( 3x + 6 = 0 \Rightarrow x = -2 \). So B = (-2, 0).

Next, we solve equations (ii) and (iii) to find vertex C:
Multiplying (iii) by 2: \( 2x - 6y - 12 = 0 \) ...(iv)

Subtracting (iv) from (ii): \( 2x - 5y + 4 - 2x + 6y + 12 = 0 \Rightarrow y + 16 = 0 \Rightarrow y = -16 \)

Substituting \( y = -16 \) into (ii): \( 2x + 80 + 4 = 0 \Rightarrow x = -42 \). So C = (-42, -16).

Finally, we solve equations (iii) and (i) to find vertex A:
Multiplying (iii) by 3: \( 3x - 9y - 18 = 0 \) ...(v)

Subtracting (v) from (i): \( 3x + 2y + 6 - 3x + 9y + 18 = 0 \Rightarrow 11y + 24 = 0 \Rightarrow y = -\frac{24}{11} \)

Substituting into (iii): \( x - 3(-\frac{24}{11}) - 6 = 0 \Rightarrow x + \frac{72}{11} - 6 = 0 \Rightarrow x = -\frac{6}{11} \). So A = \( \left(-\frac{6}{11}, -\frac{24}{11}\right) \).

The vertices are A = \( \left(-\frac{6}{11}, -\frac{24}{11}\right) \), B = (-2, 0), and C = (-42, -16).

Let D, E, and F be the midpoints of BC, CA, and AB respectively.

Coordinates of D (midpoint of BC):
\[ D = \left(\frac{-42 + (-2)}{2}, \frac{-16 + 0}{2}\right) = \left(\frac{-44}{2}, \frac{-16}{2}\right) = (-22, -8) \]

Coordinates of E (midpoint of CA):
\[ E = \left(\frac{-42 - \frac{6}{11}}{2}, \frac{-16 - \frac{24}{11}}{2}\right) = \left(\frac{-\frac{468 + 6}{11}}{2}, \frac{-\frac{176 + 24}{11}}{2}\right) = \left(-\frac{234}{11}, -\frac{100}{11}\right) \]

Coordinates of F (midpoint of AB):
\[ F = \left(\frac{-\frac{6}{11} - 2}{2}, \frac{-\frac{24}{11} + 0}{2}\right) = \left(\frac{-\frac{28}{11}}{2}, \frac{-\frac{24}{11}}{2}\right) = \left(-\frac{14}{11}, -\frac{12}{11}\right) \]

Now we find the equations of medians AD, BE, and CF.

Median AD: Passes through A\( \left(-\frac{6}{11}, -\frac{24}{11}\right) \) and D(-22, -8).

Slope = \( \frac{-8 - (-\frac{24}{11})}{-22 - (-\frac{6}{11})} = \frac{\frac{-88 + 24}{11}}{\frac{-242 + 6}{11}} = \frac{-64}{-236} = \frac{8}{29.5} \)

Equation: \( y + \frac{24}{11} = \frac{8}{29.5}\left(x + \frac{6}{11}\right) \) which simplifies to \( 8x - 29.5y + 64 = 0 \) or \( 16x - 59y + 128 = 0 \).

Median BE: Passes through B(-2, 0) and E\( \left(-\frac{234}{11}, -\frac{100}{11}\right) \).

Slope = \( \frac{-\frac{100}{11} - 0}{-\frac{234}{11} - (-2)} = \frac{-\frac{100}{11}}{\frac{-234 + 22}{11}} = \frac{-100}{-212} = \frac{25}{53} \)

Equation: \( y - 0 = \frac{25}{53}(x + 2) \) which simplifies to \( 25x - 53y + 50 = 0 \).

Median CF: Passes through C(-42, -16) and F\( \left(-\frac{14}{11}, -\frac{12}{11}\right) \).

Slope = \( \frac{-\frac{12}{11} - (-16)}{-\frac{14}{11} - (-42)} = \frac{\frac{-12 + 176}{11}}{\frac{-14 + 462}{11}} = \frac{164}{448} = \frac{41}{112} \)

Equation: \( y + 16 = \frac{41}{112}(x + 42) \) which simplifies to \( 41x - 112y + 560 = 0 \).

Exam Tip: When finding medians, always identify the vertices first by solving pairs of side equations, then calculate the midpoints of the opposite sides. The median connects a vertex to the midpoint of the opposite side. Always verify that your median equations are correctly simplified before finalizing.

Question 1. If the origin is shifted to the point (1, 2) by a translation of the axes, find the new coordinates of the point (3, -4).
Answer: Let the new origin be (h, k) = (1, 2) and (x, y) = (3, -4). Let the new coordinates be (X, Y). We apply the transformation formula: x = X + h and y = Y + k. Substituting the values: 3 = X + 1 and - 4 = Y + 2. Solving these gives X = 2 and Y = - 6. Therefore, the new coordinates are (2, -6).

Exam Tip: Always remember the transformation formula x = X + h and y = Y + k when shifting the origin, then substitute the given values carefully to find the new coordinates.

Question 2. If the origin is shifted to the point (-3, -2) by a translation of the axes, find the new coordinates of the point (3, -5).
Answer: Let the new origin be (h, k) = (-3, -2) and (x, y) = (3, -5). Let the new coordinates be (X, Y). We use the transformation formula: x = X + h and y = Y + k. Substituting: 3 = X - 3 and - 5 = Y - 2. Solving yields X = 6 and Y = - 3. Thus, the new coordinates are (6, -3).

Exam Tip: Watch the signs carefully when the new origin has negative coordinates - ensure you add or subtract correctly in the transformation equations.

Question 3. If the origin is shifted to the point (0, -2) by a translation of the axes, the coordinates of a point become (3, 2). Find the original coordinates of the point.
Answer: Let the new origin be (h, k) = (0, -2) and the new coordinates be (X, Y) = (3, 2). Let the original coordinates be (x, y). We use the transformation formula: x = X + h and y = Y + k. Substituting: x = 3 + 0 = 3 and y = 2 + (-2) = 0. Therefore, the original coordinates are (3, 0).

Exam Tip: When finding original coordinates from new coordinates, always apply x = X + h and y = Y + k directly without rearranging the formula.

Question 4. If the origin is shifted to the point (2, -1) by a translation of the axes, the coordinates of a point become (-3, 5). Find the original coordinates of the point.
Answer: Let the new origin be (h, k) = (2, -1) and the new coordinates be (X, Y) = (-3, 5). Let the original coordinates be (x, y). Using the transformation formula: x = X + h and y = Y + k. Substituting: x = -3 + 2 = -1 and y = 5 + (-1) = 4. Therefore, the original coordinates are (-1, 4).

Exam Tip: Always verify your answer by checking that the transformation formula is applied correctly in the forward direction.

Question 5. At what point must the origin be shifted, if the coordinates of a point (-4, 2) become (3, -2)?
Answer: Let (h, k) denote the point to which the origin is shifted. Given: x = -4, y = 2, X = 3, and Y = -2. Using the transformation formula x = X + h and y = Y + k, we have: -4 = 3 + h and 2 = -2 + k. Solving these: h = -7 and k = 4. Therefore, the origin must be shifted to (-7, 4).

Exam Tip: When finding the shift point, rearrange the transformation formula as h = x - X and k = y - Y to quickly find the required shift.

Question 6. Find what the given equation becomes when the origin is shifted to the point (1, 1). x² + xy - 3x - y + 2 = 0
Answer: Let the new origin be (h, k) = (1, 1). The transformation formula gives: x = X + 1 and y = Y + 1. Substituting into the given equation: (X + 1)² + (X + 1)(Y + 1) - 3(X + 1) - (Y + 1) + 2 = 0. Expanding: X² + 1 + 2X + XY + X + Y + 1 - 3X - 3 - Y - 1 + 2 = 0. Simplifying: X² + 2X + XY - 2X - 1 = 0, which gives X² + XY = 0. Therefore, the transformed equation is X² + XY = 0.

Exam Tip: Carefully expand all squared and product terms, then collect like terms to avoid errors when transforming equations.

Question 7. Find what the given equation becomes when the origin is shifted to the point (1, 1). xy - y² - x + y = 0
Answer: Let the new origin be (h, k) = (1, 1). Using the transformation formula: x = X + 1 and y = Y + 1. Substituting into the given equation: (X + 1)(Y + 1) - (Y + 1)² - (X + 1) + (Y + 1) = 0. Expanding: XY + X + Y + 1 - Y² - 1 - 2Y - X - 1 + Y + 1 = 0. Simplifying: XY - Y² = 0. Therefore, the transformed equation is XY - Y² = 0.

Exam Tip: After expanding, systematically group and cancel terms with opposite signs to reach the simplified form more easily.

Question 8. Find what the given equation becomes when the origin is shifted to the point (1, 1). x² - y² - 2x + 2y = 0
Answer: Let the new origin be (h, k) = (1, 1). Using the transformation formula: x = X + 1 and y = Y + 1. Substituting: (X + 1)² - (Y + 1)² - 2(X + 1) + 2(Y + 1) = 0. Expanding: X² + 1 + 2X - Y² - 1 - 2Y - 2X - 2 + 2Y + 2 = 0. After cancellation: X² - Y² = 0. Therefore, the transformed equation is X² - Y² = 0.

Exam Tip: Recognize that linear terms often cancel during transformation, leaving only the highest-degree terms - use this to check your algebraic work.

Question 9. Find what the given equation becomes when the origin is shifted to the point (1, 1). xy - x - y + 1 = 0
Answer: Let the new origin be (h, k) = (1, 1). Using the transformation formula: x = X + 1 and y = Y + 1. Substituting: (X + 1)(Y + 1) - (X + 1) - (Y + 1) + 1 = 0. Expanding: XY + X + Y + 1 - X - 1 - Y - 1 + 1 = 0. Simplifying: XY = 0. Therefore, the transformed equation is XY = 0.

Exam Tip: Notice when a transformation produces a very simple form like XY = 0 - this often indicates the original point (1, 1) is special relative to the curve (such as a center of symmetry).

Question 10. Transform the equation 2x² + y² - 4x + 4y = 0 to parallel axes when the origin is shifted to the point (1, -2).
Answer: Let the new origin be (h, k) = (1, -2). The transformation formula gives: x = X + 1 and y = Y - 2. Substituting into the given equation: 2(X + 1)² + (Y - 2)² - 4(X + 1) + 4(Y - 2) = 0. Expanding: 2(X² + 1 + 2X) + (Y² + 4 - 4Y) - 4X - 4 + 4Y - 8 = 0. Simplifying: 2X² + 2 + 4X + Y² + 4 - 4Y - 4X + 4Y - 12 = 0, which reduces to 2X² + Y² - 6 = 0. Therefore, the transformed equation is 2X² + Y² = 6.

Exam Tip: When all linear terms cancel after transformation, you have identified a "canonical" form that reveals the geometric structure of the curve.

Exercise 20K

Question 1. Find the equation of the line drawn through the point of intersection of the lines x - 2y + 3 = 0 and 2x - 3y + 4 = 0 and passing through the point (4, -5).
Answer: Let the two given lines intersect at point P(x₁, y₁), which satisfies both equations: x - 2y + 3 = 0 and 2x - 3y + 4 = 0. Multiplying the first equation by 2: 2x - 4y + 6 = 0. Subtracting from the second equation: y - 2 = 0, giving y = 2. Substituting back into the first equation: x - 4 + 3 = 0, so x = 1. The intersection point is P(1, 2). Next, find the slope of the line joining (1, 2) and (4, -5): m = (-5 - 2)/(4 - 1) = -7/3. Using point-slope form with point (4, -5): y + 5 = (-7/3)(x - 4). Multiplying by 3: 3y + 15 = -7x + 28. Rearranging: 7x + 3y - 13 = 0. Therefore, the equation of the required line is 7x + 3y - 13 = 0.

Exam Tip: Always find the intersection point first by solving the two given line equations simultaneously, then use this point along with the additional constraint to determine the final line equation.

Question 2. Find the equation of the line drawn through the point of intersection of the lines x - y = 7 and 2x + y = 2 and passing through the origin.
Answer: Let the two given lines intersect at point P(x₁, y₁) satisfying: x - y = 7 and 2x + y = 2. Multiplying the first equation by 2: 2x - 2y = 14. Subtracting from the second: -3y = 12, so y = -4. Substituting into the first: x + 4 = 7, giving x = 3. The intersection point is P(3, -4). Now find the slope of the line joining (3, -4) and the origin (0, 0): m = (0 - (-4))/(0 - 3) = -4/3. Using the origin as the reference point with this slope: y - 0 = (-4/3)(x - 0). Simplifying: 3y = -4x, or 4x + 3y = 0. Therefore, the equation of the required line is 4x + 3y = 0.

Exam Tip: When a line passes through the origin, remember that the y-intercept is zero, which simplifies the final equation form significantly.

Question 3. Find the equation of the line drawn through the point of intersection of the lines x + y = 9 and 2x - 3y + 7 = 0 and whose slope is -2/3.
Answer: Let the two given lines intersect at point P(x₁, y₁) satisfying: x + y = 9 and 2x - 3y + 7 = 0. Multiplying the first by 2: 2x + 2y = 18. Subtracting from the second: -5y + 25 = 0, so y = 5. Substituting back: x + 5 = 9, giving x = 4. The intersection point is P(4, 5). Now find the equation of a line passing through (4, 5) with slope m = -2/3. Using point-slope form: y - 5 = (-2/3)(x - 4). Multiplying by 3: 3y - 15 = -2x + 8. Rearranging: 2x + 3y - 23 = 0. Therefore, the equation of the required line is 2x + 3y - 23 = 0.

Exam Tip: When the slope is explicitly given, skip the slope calculation step and go directly to the point-slope formula using the intersection point and the given slope value.

Question 4. Find the equation of the line drawn through the point of intersection of the lines x - y = 1 and 2x - 3y + 1 = 0 and which is parallel to the line 3x + 4y = 12.
Answer: Assume the two given lines meet at a point P(x₁, y₁). Then (x₁, y₁) must satisfy each of the given equations.

\( x - y = 1 \) ...(i)
\( 2x - 3y + 1 = 0 \) ...(ii)

To locate the intersection point of equations (i) and (ii):

Multiply equation (i) by 2:
\( 2x - 2y = 2 \)
or \( 2x - 2y - 2 = 0 \) ...(iii)

Subtract equation (iii) from equation (ii):
\( 2x - 3y + 1 - 2x + 2y + 2 = 0 \)
\( \implies -y + 3 = 0 \)
\( \implies y = 3 \)

Substitute y = 3 into equation (i):
\( x - 3 = 1 \)
\( \implies x = 4 \)

Therefore, the intersection point P(x₁, y₁) is (4, 3).

Next, determine the slope of the line \( 3x + 4y = 12 \). The slope formula for an equation is \( m = -\frac{a}{b} \), so:
\( m = -\frac{3}{4} \)

A line parallel to this one has the same slope: \( -\frac{3}{4} \)

Now find the equation of the line passing through (4, 3) with slope \( -\frac{3}{4} \):
\( y - y_1 = m(x - x_1) \)
\( \implies y - 3 = -\frac{3}{4}(x - 4) \)
\( \implies y - 3 = -3x + 12 \)
\( \implies 4y - 12 = -3x + 12 \)
\( \implies 3x + 4y - 24 = 0 \)

Exam Tip: Always identify the slopes correctly - parallel lines share the same slope, and use the point-slope form to build the final equation quickly.

Question 5. Find the equation of the line through the intersection of the lines 5x - 3y = 1 and 2x + 3y = 23 and which is perpendicular to the line 5x - 3y = 1.
Answer: Assume the two given lines meet at a point P(x₁, y₁). Then (x₁, y₁) must satisfy each of the given equations.

\( 5x - 3y = 1 \) ...(i)
\( 2x + 3y = 23 \) ...(ii)

To locate the intersection point of equations (i) and (ii):

Add equations (i) and (ii):
\( 5x - 3y + 2x + 3y = 1 + 23 \)
\( \implies 7x = 24 \)
\( \implies x = \frac{24}{7} \)

Substitute \( x = \frac{24}{7} \) into equation (i):
\( 5 \left( \frac{24}{7} \right) - 3y = 1 \)
\( \implies \frac{120}{7} - 3y = 1 \)
\( \implies -3y = 1 - \frac{120}{7} \)
\( \implies -3y = \frac{7 - 120}{7} \)
\( \implies -3y = -\frac{113}{7} \)
\( \implies y = \frac{113}{21} \)

Therefore, the intersection point P(x₁, y₁) is \( \left( \frac{24}{7}, \frac{113}{21} \right) \).

When two lines are perpendicular, the product of their slopes equals - 1:
\( m_1 \times m_2 = -1 \)

For the line \( 5x - 3y = 1 \), the slope is \( \frac{5}{3} \). Therefore:
\( \frac{5}{3} \times m_{\text{perp}} = -1 \)
\( \implies m_{\text{perp}} = -\frac{3}{5} \)

Find the equation of the line passing through \( \left( \frac{24}{7}, \frac{113}{21} \right) \) with slope \( -\frac{3}{5} \):
\( y - y_1 = m(x - x_1) \)
\( \implies y - \frac{113}{21} = -\frac{3}{5} \left( x - \frac{24}{7} \right) \)
\( \implies 5y - 5 \times \frac{113}{21} = -3x + \frac{24}{7} \)
\( \implies 5y - \frac{565}{21} = -3x + \frac{72}{7} \)
\( \implies 3x + 5y - \frac{565}{21} - \frac{72}{7} = 0 \)
\( \implies 3x + 5y - \frac{565}{21} - \frac{216}{21} = 0 \)
\( \implies \frac{63x + 105y - 565 - 216}{21} = 0 \)
\( \implies 63x + 105y - 781 = 0 \)

Exam Tip: For perpendicular lines, remember the slope relationship m₁ × m₂ = - 1; clear fractions early to avoid arithmetic errors in final steps.

Question 6. Find the equation of the line through the intersection of the lines 2x - 3y = 0 and 4x - 5y = 2 and which is perpendicular to the line x + 2y + 1 = 0.
Answer: Assume the two given lines meet at a point P(x₁, y₁). Then (x₁, y₁) must satisfy each of the given equations.

\( 2x - 3y = 0 \) ...(i)
\( 4x - 5y = 2 \) ...(ii)

To locate the intersection point of equations (i) and (ii):

Multiply equation (i) by 2:
\( 4x - 6y = 0 \) ...(iii)

Subtract equation (iii) from equation (ii):
\( 4x - 5y - 4x + 6y = 2 - 0 \)
\( \implies y = 2 \)

Substitute y = 2 into equation (i):
\( 2x - 3(2) = 0 \)
\( \implies 2x = 6 \)
\( \implies x = 3 \)

Therefore, the intersection point P(x₁, y₁) is (3, 2).

When two lines are perpendicular, the product of their slopes equals - 1. The line \( x + 2y + 1 = 0 \) can be rewritten as \( 2y = -x - 1 \), giving slope \( m_1 = -\frac{1}{2} \). Therefore:
\( -\frac{1}{2} \times m_{\text{perp}} = -1 \)
\( \implies m_{\text{perp}} = 2 \)

Find the equation of the line passing through (3, 2) with slope 2:
\( y - y_1 = m(x - x_1) \)
\( \implies y - 2 = 2(x - 3) \)
\( \implies y - 2 = 2x - 6 \)
\( \implies 2x - y - 4 = 0 \)

Exam Tip: Always convert the given line to slope-intercept form first to identify its slope accurately, then apply the perpendicularity condition.

Question 7. Find the equation of the line through the intersection of the lines x - 7y + 5 = 0 and 3x + y - 7 = 0 and which is parallel to x-axis.
Answer: Assume the two given lines meet at a point P(x₁, y₁). Then (x₁, y₁) must satisfy each of the given equations.

\( x - 7y + 5 = 0 \) ...(i)
\( 3x + y - 7 = 0 \) ...(ii)

To locate the intersection point of equations (i) and (ii):

Multiply equation (i) by 3:
\( 3x - 21y + 15 = 0 \) ...(iii)

Subtract equation (iii) from equation (ii):
\( 3x + y - 7 - 3x + 21y - 15 = 0 \)
\( \implies 22y - 22 = 0 \)
\( \implies y = 1 \)

Substitute y = 1 into equation (i):
\( x - 7(1) + 5 = 0 \)
\( \implies x - 2 = 0 \)
\( \implies x = 2 \)

Therefore, the intersection point P(x₁, y₁) is (2, 1).

Any line parallel to the x-axis has the form \( y = b \), where b is a constant. Since the required line passes through (2, 1), the y-coordinate at this point must be 1. Therefore:
\( y = 1 \)

Exam Tip: Lines parallel to the x-axis are always horizontal with equation y = k; lines parallel to the y-axis are vertical with equation x = k. Identify which axis the problem specifies.

Question 8. Find the equation of the line through the intersection of the lines 2x - 3y + 1 = 0 and x + y - 2 = 0 and drawn parallel to y-axis.
Answer: Assume the two given lines meet at a point P(x₁, y₁). Then (x₁, y₁) must satisfy each of the given equations.

\( 2x - 3y + 1 = 0 \) ...(i)
\( x + y - 2 = 0 \) ...(ii)

To locate the intersection point of equations (i) and (ii):

Multiply equation (ii) by 2:
\( 2x + 2y - 4 = 0 \) ...(iii)

Subtract equation (iii) from equation (i):
\( 2x - 3y + 1 - 2x - 2y + 4 = 0 \)
\( \implies -5y + 5 = 0 \)
\( \implies y = 1 \)

Substitute y = 1 into equation (ii):
\( x + 1 - 2 = 0 \)
\( \implies x = 1 \)

Therefore, the intersection point P(x₁, y₁) is (1, 1).

Any line parallel to the y-axis has the form \( x = a \), where a is a constant. Since the required line passes through (1, 1), the x-coordinate at this point must be 1. Therefore:
\( x = 1 \)

Exam Tip: Vertical lines (parallel to y-axis) cannot be written in slope-intercept form because the slope is undefined; always use x = k form instead.

Question 9. Find the equation of the line through the intersection of the lines 2x + 3y - 2 = 0 and x - 2y + 1 = 0 and having x-intercept equal to 3.
Answer: Assume the two given lines meet at a point P(x₁, y₁). Then (x₁, y₁) must satisfy each of the given equations.

\( 2x + 3y - 2 = 0 \) ...(i)
\( x - 2y + 1 = 0 \) ...(ii)

To locate the intersection point of equations (i) and (ii):

Multiply equation (ii) by 2:
\( 2x - 4y + 2 = 0 \) ...(iii)

Subtract equation (iii) from equation (i):
\( 2x + 3y - 2 - 2x + 4y - 2 = 0 \)
\( \implies 7y - 4 = 0 \)
\( \implies y = \frac{4}{7} \)

Substitute \( y = \frac{4}{7} \) into equation (ii):
\( x - 2 \left( \frac{4}{7} \right) + 1 = 0 \)
\( \implies x - \frac{8}{7} + 1 = 0 \)
\( \implies x + \frac{-8 + 7}{7} = 0 \)
\( \implies x = \frac{1}{7} \)

Therefore, the intersection point P(x₁, y₁) is \( \left( \frac{1}{7}, \frac{4}{7} \right) \).

The x-intercept of a line is the x-coordinate when y = 0. If the x-intercept is 3, the line passes through the point (3, 0). Using the two-point form with \( \left( \frac{1}{7}, \frac{4}{7} \right) \) and (3, 0):

Slope: \( m = \frac{0 - \frac{4}{7}}{3 - \frac{1}{7}} = \frac{-\frac{4}{7}}{\frac{21 - 1}{7}} = \frac{-\frac{4}{7}}{\frac{20}{7}} = -\frac{4}{20} = -\frac{1}{5} \)

Equation using point (3, 0):
\( y - 0 = -\frac{1}{5}(x - 3) \)
\( \implies y = -\frac{1}{5}x + \frac{3}{5} \)
\( \implies 5y = -x + 3 \)
\( \implies x + 5y - 3 = 0 \)

Exam Tip: When the x-intercept is given, use (x-intercept, 0) as one of your reference points; this directly fixes one point on the required line and simplifies calculation of the slope.

Question 10. Find the equation of the line passing through the intersection of the lines 3x - 4y + 1 = 0 and 5x + y - 1 = 0 and which cuts off equal intercepts from the axes.
Answer: Let's assume the two given lines meet at a point P(x1, y1). Then (x1, y1) satisfies both given equations.

3x - 4y + 1 = 0 ...(i)

5x + y - 1 = 0 ...(ii)

To find where eq. (i) and (ii) intersect, we multiply eq. (ii) by 4:

20x + 4y - 4 = 0 ...(iii)

Adding eq. (iii) and (i):

20x + 4y - 4 + 3x - 4y + 1 = 0

\( \Rightarrow \) 23x - 3 = 0

\( \Rightarrow \) 23x = 3

\( \Rightarrow x = \frac{3}{23} \)

Substituting this value into eq. (ii):

\( 5\left(\frac{3}{23}\right) + y - 1 = 0 \)

\( \Rightarrow \frac{15}{23} + y - 1 = 0 \)

\( \Rightarrow y = 1 - \frac{15}{23} \)

\( \Rightarrow y = \frac{23 - 15}{23} \)

\( \Rightarrow y = \frac{8}{23} \)

So the intersection point P(x1, y1) is \( \left(\frac{3}{23}, \frac{8}{23}\right) \)

The intercept form of a line's equation is:

\( \frac{x}{a} + \frac{y}{b} = 1 \)

where a and b represent the intercepts on the axes. Since equal intercepts are cut off, we have a = b.

\( \Rightarrow \frac{x}{a} + \frac{y}{a} = 1 \)

\( \Rightarrow \frac{x + y}{a} = 1 \)

\( \Rightarrow x + y = a \) ...(i)

Since this line passes through \( \left(\frac{3}{23}, \frac{8}{23}\right) \), we substitute:

\( \frac{3}{23} + \frac{8}{23} = a \)

\( \Rightarrow \frac{11}{23} = a \)

Putting this value into eq. (i):

\( x + y = \frac{11}{23} \)

\( \Rightarrow 23x + 23y = 11 \)

Therefore, the required equation of the line is 23x + 23y = 11
In simple words: We first locate where the two given lines cross each other. Then, using the fact that the line we're looking for cuts equal intercepts on both axes, we write its equation in intercept form. We substitute the intersection point into this equation to find the intercept value, giving us our final answer.

Exam Tip: Always verify your intersection point by substituting it back into both original equations. For equal intercepts, remember that a = b in the intercept form, which simplifies the equation significantly.