Maharashtra Board Class 10 Maths Chapter 5 Co ordinate Geometry Set 5.1 Solutions

Question 1. Find the distance between each of the following pairs of points.
(i) A (2, 3), B (4,1)
(ii) P (-5, 7), Q (-1, 3)
(iii) R (0, -3), S \( (0,\frac{5}{2}) \)
(iv) L (5, -8), M (-7, -3)
(v) T (-3, 6), R (9, -10)
(vi) W \( (-\frac{7}{2}, 4) \), X(11, 4)
Answer: Solution:
(i) Let A \( (x_1, y_1) \) and B \( (x_2, y_2) \) be the given points.
\( \therefore x_1 = 2, y_1 = 3, x_2 = 4, y_2 = 1 \)
By distance formula,
\( d(A, B) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{(4-2)^2+(1-3)^2} \)
\( = \sqrt{2^2+(-2)^2} \)
\( = \sqrt{4+4} \)
\( = \sqrt{8} \)
\( \therefore d(A, B) = 2\sqrt{2} \) units
\( \therefore \) The distance between the points A and B is \( 2\sqrt{2} \) units.
(ii) Let P \( (x_1, y_1) \) and Q \( (x_2, y_2) \) be the given points.
\( \therefore x_1 = -5, y_1 = 7, x_2 = -1, y_2 = 3 \)
By distance formula,
\( d(P, Q) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{[-1-(-5)]^2+(3-7)^2} \)
\( = \sqrt{(-1+5)^2+(-4)^2} \)
\( = \sqrt{4^2+(-4)^2} \)
\( = \sqrt{16+16} \)
\( = \sqrt{32} \)
\( \therefore d(P, Q) = 4\sqrt{2} \) units
\( \therefore \) The distance between the points P and Q is \( 4\sqrt{2} \) units.
(iii) Let R \( (x_1, y_1) \) and S \( (x_2, y_2) \) be the given points.
\( \therefore x_1 = 0, y_1 = -3, x_2 = 0, y_2 = \frac{5}{2} \)
By distance formula,
\( d(R, S) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{(0-0)^2+(\frac{5}{2}-(-3))^2} \)
\( = \sqrt{(\frac{5}{2}+3)^2} \)
\( = \sqrt{(\frac{5+6}{2})^2} \)
\( = \sqrt{(\frac{11}{2})^2} \)
\( = \frac{11}{2} \)
\( \therefore d(R, S) = \frac{11}{2} \) units
\( \therefore \) The distance between the points R and S is \( \frac{11}{2} \) units.
(iv) Let L \( (x_1, y_1) \) and M \( (x_2, y_2) \) be the given points.
\( \therefore x_1 = 5, y_1 = -8, x_2 = -7, y_2 = -3 \)
By distance formula,
\( d(L, M) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{(-7-5)^2 +[-3-(-8)]^2} \)
\( = \sqrt{(-7-5)^2+(-3+8)^2} \)
\( = \sqrt{(-12)^2 + (5)^2} \)
\( = \sqrt{144+25} \)
\( = \sqrt{169} \)
\( \therefore d(L, M) = 13 \) units
\( \therefore \) The distance between the points L and M is 13 units.
(v) Let T \( (x_1,y_1) \) and R \( (x_2, y_2) \) be the given points.
\( \therefore x_1 = -3, y_1 = 6, x_2 = 9, y_2 = -10 \)
By distance formula,
\( d(T, R) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{[9-(-3)]^2+(-10-6)^2} \)
\( = \sqrt{(9+3)^2+(-10-6)^2} \)
\( = \sqrt{12^2 + (-16)^2} \)
\( = \sqrt{144+256} \)
\( = \sqrt{400} \)
\( \therefore d(T, R) = 20 \) units
\( \therefore \) The distance between the points T and R 20 units.
(vi) Let W \( (x_1, y_1) \) and X \( (x_2, y_2) \) be the given points.
\( x_1 = -\frac{7}{2}, y_1 = 4, x_2 = 11, y_2 = 4 \)
By distance formula,
\( d(W, X) = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \)
\( = \sqrt{[11-(-\frac{7}{2})]^2+(4-4)^2} \)
\( = \sqrt{(11+\frac{7}{2})^2} \)
\( = \sqrt{(\frac{22+7}{2})^2} \)
\( = \sqrt{(\frac{29}{2})^2} \)
\( = \frac{29}{2} \)
\( \therefore d(W, X) = \frac{29}{2} \) units
\( \therefore \) The distance between the points W and X is \( \frac{29}{2} \) units.
In simple words: This question uses the distance formula to calculate the length between several pairs of points in a coordinate plane, applying it systematically for different coordinate types including fractions and negative numbers.

🎯 Exam Tip: Ensure precise application of the distance formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \), paying close attention to signs of coordinates and fractional arithmetic for full marks.

Question 2. Determine whether the points are collinear.
(i) A (1, -3), B (2, -5), C (-4, 7)
(ii) L (-2, 3), M (1, -3), N (5, 4)
(iii) R (0, 3), D (2, 1), S (3, -1)
(iv) P (-2, 3), Q (1, 2), R (4, 1)
Answer: Solution:
(i) By distance formula,
\( d(A, B) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{(2-1)^2+[-5-(-3)]^2} \)
\( = \sqrt{(2-1)^2+(-5+3)^2} \)
\( = \sqrt{1^2 + (-2)^2} \)
\( = \sqrt{1+4} \)
\( = \sqrt{5} \) ...(i)
\( d(B, C) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{(-4-2)^2 +[7-(-5)]^2} \)
\( = \sqrt{(-4-2)^2+(7+5)^2} \)
\( = \sqrt{(-6)^2 +12^2} \)
\( = \sqrt{36+144} \)
\( = \sqrt{180} \)
\( \therefore d(B, C) = 6\sqrt{5} \) ...(ii)
\( d(A, C) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{(-4-1)^2 +[7-(-3)]^2} \)
\( = \sqrt{(-4-1)^2+(7+3)^2} \)
\( = \sqrt{(-5)^2 +10^2} \)
\( = \sqrt{25+100} \)
\( = \sqrt{125} \)
\( = 5\sqrt{5} \) ...(iii)
On adding (i) and (iii),
\( d(A, B) + d(A, C) = \sqrt{5} + 5\sqrt{5} = 6\sqrt{5} \)
\( \therefore d(A, B) + d(A, C) = d(B, C) \) ... [From (ii)]
\( \therefore \) Points A, B and C are collinear.
(ii) By distance formula,
\( d(L, M) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{[1-(-2)]^2+(-3-3)^2} \)
\( = \sqrt{(1+2)^2+(-3-3)^2} \)
\( = \sqrt{3^2 +(-6)^2} \)
\( = \sqrt{9+36} \)
\( = \sqrt{45} \)
\( \therefore d(L, M) = 3\sqrt{5} \) ...(i)
\( d(M, N) = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \)
\( = \sqrt{(5-1)^2 +[4-(-3)]^2} \)
\( = \sqrt{(5-1)^2+(4+3)^2} \)
\( = \sqrt{4^2+7^2} \)
\( = \sqrt{16+49} \)
\( = \sqrt{65} \)
\( \therefore d(M, N) = \sqrt{65} \) ...(ii)
\( d(L, N) = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \)
\( = \sqrt{[5-(-2)]^2+(4-3)^2} \)
\( = \sqrt{(5+2)^2+(4-3)^2} \)
\( = \sqrt{7^2 + 1^2} \)
\( = \sqrt{49+1} \)
\( = \sqrt{50} \)
\( \therefore d(L, N) = 5\sqrt{2} \) ...(iii)
On adding (i) and (iii),
\( d(L, M) + d(L, N) = 3\sqrt{5} + 5\sqrt{2} \neq \sqrt{65} \)
\( \therefore d(L, M) + d(L, N) \neq d(M, N) \) ... [From (ii)]
\( \therefore \) Points L, M and N are not collinear.
(iii) By distance formula,
\( d (R, D) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{(2-0)^2+(1-3)^2} \)
\( = \sqrt{2^2+(-2)^2} \)
\( = \sqrt{4+4} \)
\( = \sqrt{8} \) ...(i)
\( d(D, S) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{(3-2)^2+(-1-1)^2} \)
\( = \sqrt{1^2 + (-2)^2} \)
\( = \sqrt{1+4} \)
\( = \sqrt{5} \) ...(ii)
\( d(R, S) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{(3-0)^2+(-1-3)^2} \)
\( = \sqrt{3^2 + (-4)^2} \)
\( = \sqrt{9+16} \)
\( = \sqrt{25} \)
\( \therefore d(R, S) = 5 \) ...(iii)
On adding (i) and (ii),
\( \therefore d(R, D) + d(D, S) = \sqrt{8} + \sqrt{5} \neq 5 \)
\( \therefore d(R, D) + d(D, S) \neq d(R, S) \) ... [From (iii)]
\( \therefore \) Points R, D and S are not collinear.
(iv) By distance formula,
\( d(P, Q) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{[1-(-2)]^2+(2-3)^2} \)
\( = \sqrt{(1+2)^2+(-3-3)^2} \)
\( = \sqrt{3^2+(-1)^2} \)
\( = \sqrt{9+1} \)
\( = \sqrt{10} \) ...(i)
\( d(Q, R) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{(4-1)^2+(1-2)^2} \)
\( = \sqrt{3^2 + (-1)^2} \)
\( = \sqrt{9+1} \)
\( = \sqrt{10} \) ...(ii)
\( d(P, R) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( = \sqrt{[4-(-2)]^2+(1-3)^2} \)
\( = \sqrt{(4+2)^2+(1-3)^2} \)
\( = \sqrt{6^2+(-2)^2} \)
\( = \sqrt{36+4} \)
\( = \sqrt{40} \)
\( \therefore d(P, R) = 2\sqrt{10} \) ...(iii)
On adding (i) and (ii),
\( d(P, Q) + d(Q, R) = \sqrt{10} + \sqrt{10} = 2\sqrt{10} \)
\( \therefore d(P, Q) + d(Q, R) = d(P, R) \) ... [From (iii)]
\( \therefore \) Points P, Q and R are collinear.
In simple words: Points are collinear if the sum of the distances between two pairs of points equals the distance of the third pair. If a + b = c, then the points lie on the same line.

🎯 Exam Tip: To prove collinearity, calculate all three possible distances between the given points. If the sum of any two distances equals the third distance, the points are collinear.

Question 3. Find the point on the X-axis which is equidistant from A (-3,4) and B (1, -4).
Answer: Solution:
Let point C be on the X-axis which is equidistant from points A and B.
Point C lies on X-axis.
\( \therefore \) its y co-ordinate is 0.
Let C = (x, 0)
C is equidistant from points A and B.
\( \therefore \) AC = BC
\( \sqrt{[x - (-3)]^2 + (0 - 4)^2} = \sqrt{(x - 1)^2 + [0 - (-4)]^2} \) ... [By distance formula]
\( [x-(-3)]^2 + (0-4)^2 = (x-1)^2 + [0-(-4)]^2 \) ... [Squaring both sides]
\( \therefore (x + 3)^2 + (-4)^2 = (x-1)^2 + 4^2 \)
\( \therefore x^2 + 6x + 9 + 16 = x^2 - 2x + 1 + 16 \)
\( \therefore 8x = - 8 \)
\( \therefore x = -\frac{8}{8} = -1 \)
\( \therefore \) The point on X-axis which is equidistant from points A and B is (-1,0).
In simple words: We find a point (x, 0) on the X-axis that is the same distance from two given points by setting the distances equal using the distance formula and solving for x.

🎯 Exam Tip: Remember that any point on the X-axis has a y-coordinate of 0. Equating the squared distances avoids square roots and simplifies calculations.

Question 4. Verify that points P (-2, 2), Q (2, 2) and R (2, 7) are vertices of a right angled triangle.
Answer: Solution:
Distance between two points
\( = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
By distance formula,
\( PQ = \sqrt{[2-(-2)]^2 +(2-2)^2} \)
\( = \sqrt{(2+2)^2+(0)^2} \)
\( = \sqrt{(4)^2} \)
\( = 4 \) ...(i)
\( QR = \sqrt{(2-2)^2+(7-2)^2} \)
\( = \sqrt{(0)^2 +(5)^2} \)
\( = \sqrt{(5)^2} \)
\( = 5 \) ...(ii)
\( PR = \sqrt{[2-(-2)]^2+(7-2)^2} \)
\( = \sqrt{(2+2)^2+(5)^2} \)
\( = \sqrt{(4)^2+(5)^2} \)
\( = \sqrt{16+25} \)
\( = \sqrt{41} \) ...(iii)
Now, \( PR^2 = (\sqrt{41})^2 = 41 \)
Consider, \( PQ^2 + QR^2 = 4^2 + 5^2 = 16 + 25 = 41 \) [From (i) and (ii)]
\( \therefore PR^2 = PQ^2 + QR^2 \) ... [From (iii)]
\( \therefore \triangle PQR \) is a right angled triangle. ... [Converse of Pythagoras theorem]
\( \therefore \) Points P, Q and R are the vertices of a right angled triangle.
In simple words: To verify if points form a right-angled triangle, calculate the square of the lengths of all three sides using the distance formula. If the square of the longest side equals the sum of the squares of the other two sides, it's a right-angled triangle, based on the Pythagorean theorem.

🎯 Exam Tip: Always calculate the squares of the distances and then check if the Pythagorean relationship \( (a^2 + b^2 = c^2) \) holds true. Clearly state the application of the converse of the Pythagoras theorem.

Question 5. Show that points P (2, -2), Q (7, 3), R (11, -1) and S (6, -6) are vertices of a parallelogram.
Answer: Proof:
Distance between two points
\( = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
By distance formula,
\( PQ = \sqrt{(7-2)^2 +[3-(-2)]^2} \)
\( = \sqrt{(7-2)^2+(3+2)^2} \)
\( = \sqrt{5^2+5^2} \)
\( = \sqrt{25+25} \)
\( = \sqrt{50} \) ...(i)
\( QR = \sqrt{(11-7)^2+(-1-3)^2} \)
\( = \sqrt{4^2 +(-4)^2} \)
\( = \sqrt{16+16} \)
\( = \sqrt{32} \) ...(ii)
\( RS = \sqrt{(6-11)^2 +[-6-(-1)]^2} \)
\( = \sqrt{(6-11)^2 + (-6 + 1)^2} \)
\( = \sqrt{(-5)^2 + (-5)^2} \)
\( = \sqrt{25 +25} \)
\( = \sqrt{50} \) ...(iii)
\( PS = \sqrt{(6-2)^2+[-6-(-2)]^2} \)
\( = \sqrt{(6-2)^2 + (-6+2)^2} \)
\( = \sqrt{4^2 +(-4)^2} \)
\( = \sqrt{16+16} \)
\( = \sqrt{32} \) ...(iv)
PQ = RS ... [From (i) and (iii)]
QR = PS ... [From (ii) and (iv)]
A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.
\( \therefore \Box PQRS \) is a parallelogram.
\( \therefore \) Points P, Q, R and S are the vertices of a parallelogram.
In simple words: To prove that a quadrilateral is a parallelogram, we calculate the lengths of all four sides using the distance formula. If both pairs of opposite sides are equal in length, then the figure is a parallelogram.

🎯 Exam Tip: For proving a parallelogram, show that opposite sides are equal in length. Clearly label distances and reference corresponding equations for a coherent proof.

Question 6. Show that points A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are vertices of rhombus ABCD.
Answer: Proof:
Distance between two points
\( = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
By distance formula,
\( AB= \sqrt{[-1-(-4)]^2+[2-(-7)]^2} \)
\( = \sqrt{(-1+4)^2 + (2+7)^2} \)
\( = \sqrt{3^2 +9^2} \)
\( = \sqrt{9+81} \)
\( = \sqrt{90} \) ...(i)
\( BC= \sqrt{[8-(-1)]^2+(5-2)^2} \)
\( = \sqrt{(8 + 1)^2 + (5-2)^2} \)
\( = \sqrt{9^2 + 3^2} \)
\( = \sqrt{81+9} \)
\( = \sqrt{90} \) ...(ii)
\( CD = \sqrt{(5-8)^2+(-4-5)^2} \)
\( = \sqrt{(-3)^2 + (-9)^2} \)
\( = \sqrt{9+81} \)
\( = \sqrt{90} \) ...(iii)
\( AD= \sqrt{[5-(-4)]^2+[-4-(-7)]^2} \)
\( = \sqrt{(5+4)^2+(-4+7)^2} \)
\( = \sqrt{9^2 + 3^2} \)
\( = \sqrt{81+9} \)
\( = \sqrt{90} \) ...(iv)
\( \therefore AB = BC = CD = AD \) ... [From (i), (ii), (iii) and (iv)]
In a quadrilateral, if all the sides are equal, then it is a rhombus.
\( \therefore \Box ABCD \) is a rhombus.
\( \therefore \) Points A, B, C and D are the vertices of rhombus ABCD.
In simple words: To demonstrate that the given points form a rhombus, calculate the lengths of all four sides of the quadrilateral using the distance formula. If all four sides are equal in length, then the quadrilateral is a rhombus.

🎯 Exam Tip: For proving a rhombus, calculate all four side lengths. Showing all sides are equal is sufficient. No need to calculate diagonals unless explicitly asked.

Question 7. Find x if distance between points L (x, 7) and M (1,15) is 10.
Answer: Solution:
\( x_1 = x, y_1 = 7, x_2 = 1, y_2 = 15 \)
By distance formula,
\( d(L, M) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( \therefore d(L, M) = \sqrt{(1-x)^2 + (15-7)^2} \)
\( \therefore 10 = \sqrt{(1-x)^2 +8^2} \)
\( \therefore 100 = (1-x)^2+64 \) ... [Squaring both sides]
\( \therefore (1-x)^2 = 100-64 \)
\( \therefore (1-x)^2 = 36 \)
\( \therefore 1-x = \pm\sqrt{36} \) ... [Taking square root of both sides]
\( \therefore 1-x = \pm 6 \)
\( \therefore 1-x = 6 \) or \( 1-x = -6 \)
\( \therefore x = - 5 \) or \( x = 7 \)
\( \therefore \) The value of x is - 5 or 7.
In simple words: Given the distance between two points, one with an unknown x-coordinate, we use the distance formula, set up an equation, square both sides to remove the square root, and solve the resulting quadratic equation for x.

🎯 Exam Tip: Remember to consider both positive and negative roots when taking the square root of a number, as \( (x)^2 = a \) yields \( x = \pm\sqrt{a} \). This often results in two possible values for the unknown coordinate.

Question 8. Show that the points A (1, 2), B (1, 6), C \( (1 + 2\sqrt{3}, 4) \) are vertices of an equilateral triangle.
Answer: Proof:
Distance between two points
\( = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
By distance formula,
\( AB = \sqrt{(1-1)^2 +(6-2)^2} \)
\( = \sqrt{0^2 + 4^2} \)
\( = \sqrt{4^2} \)
\( = 4 \) ...(i)
\( BC = \sqrt{(1+2\sqrt{3}-1)^2+(4-6)^2} \)
\( = \sqrt{(2\sqrt{3})^2+(-2)^2} \)
\( = \sqrt{12+4} \)
\( = \sqrt{16} \)
\( = 4 \) ...(ii)
\( AC = \sqrt{(1+2\sqrt{3}-1)^2+(4-2)^2} \)
\( = \sqrt{(2\sqrt{3})^2+2^2} \)
\( = \sqrt{12+4} \)
\( = \sqrt{16} \)
\( = 4 \) ...(iii)
\( \therefore AB = BC = AC \) ... [From (i), (ii) and (iii)]
\( \therefore \triangle ABC \) is an equilateral triangle.
\( \therefore \) Points A, B and C are the vertices of an equilateral triangle.
In simple words: To prove that three points form an equilateral triangle, calculate the lengths of all three sides using the distance formula. If all three side lengths are equal, then the triangle is equilateral.

🎯 Exam Tip: When dealing with coordinates involving square roots, be meticulous with squaring and combining terms. Clearly showing that all three side lengths are equal is key for proving an equilateral triangle.

Question 1. In the figure, seg AB || Y-axis and seg CB || X-axis. Co-ordinates of points A and C are given. To find AC, fill in the boxes given below. (Textbook pa. no. 102)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समन्वय तल में एक त्रिभुज ABC को दर्शाता है। बिंदु A (2, 3) पर है, बिंदु C (-2, 2) पर है, और बिंदु B (2, 2) पर है। AB रेखा Y-अक्ष के समानांतर है और CB रेखा X-अक्ष के समानांतर है, जो इंगित करती है कि बिंदु B पर एक समकोण है।
Answer: Solution:
In \( \triangle ABC, \angle B = 90^\circ \)
\( \therefore (AB)^2 + (BC)^2 = [(AC)^2] \) ... [Pythagoras theorem]
seg CB || X-axis
\( \therefore \) y co-ordinate of B = 2
seg BA || Y-axis
\( \therefore \) x co-ordinate of B = 2
\( \therefore \) co-ordinate of B is (2, 2) \( = (x_1,y_1) \)
co-ordinate of A is (2, 3) \( = (x_2,y_2) \)
Since, AB || to Y-axis,
d(A, B) = \( y_2 - y_1 \)
d(A,B) = \( 3 - 2 = 1 \)
co-ordinate of C is (-2,2) \( = (x_1,y_1) \)
co-ordinate of B is (2, 2) \( = (x_2, y_2) \)
Since, BC || to X-axis,
d(B, C) = \( x_2 - x_1 \)
d(B,C) = \( 2 - (-2) = 4 \)
\( \therefore AC^2 = 1^2 + 4^2 \) ... [From (i)]
\( = 1 + 16 = 17 \)
\( \therefore AC = \sqrt{17} \) units ... [Taking square root of both sides]
In simple words: We used the given coordinates and the fact that AB is parallel to the Y-axis and CB is parallel to the X-axis to find the coordinates of point B, and then calculated the lengths of AB and BC. Finally, we applied the Pythagorean theorem to find the length of AC.

🎯 Exam Tip: When given a diagram on a coordinate plane, identify parallel lines to axes to easily determine coordinate values. Use the Pythagorean theorem for right-angled triangles to find unknown lengths efficiently.

Question 1. Find the distance between each of the following pairs of points.
(i) A (2, 3), B (4,1)
(ii) P (-5, 7), Q (-1, 3)
(iii) R (0, -3), S \( (0,\frac{5}{2}) \)
(iv) L (5, -8), M (-7, -3)
(v) T (-3, 6), R (9, -10)
(vi) W \( (-\frac{7}{2}, 4) \), X(11, 4)
Answer:
(i) Let A (\(x_1\), \(y_1\)) and B (\(x_2\), \(y_2\)) be the given points.
\(x_1 = 2, y_1 = 3, x_2 = 4, y_2 = 1\)
By distance formula,
\(d(A, B) = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
\( = \sqrt{(4-2)^2+(1-3)^2} = \sqrt{2^2+(-2)^2}\)
\( = \sqrt{4+4} = \sqrt{8}\)
\(d(A, B) = 2\sqrt{2}\) units
The distance between the points A and B is \(2\sqrt{2}\) units.
(ii) Let P (\(x_1\), \(y_1\)) and Q (\(x_2\), \(y_2\)) be the given points.
\(x_1 = -5, y_1 = 7, x_2 = -1, y_2 = 3\)
By distance formula,
\(d(P, Q) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{[-1-(-5)]^2+(3-7)^2}\)
\( = \sqrt{(-1+5)^2+(3-7)^2}\)
\( = \sqrt{4^2+(-4)^2}\)
\( = \sqrt{16+16}\)
\( = \sqrt{32}\)
\(d(P, Q) = 4\sqrt{2}\) units
The distance between the points P and Q is \(4\sqrt{2}\) units.
(iii) Let R (\(x_1\), \(y_1\)) and S (\(x_2\), \(y_2\)) be the given points.
\(x_1 = 0, y_1 = -3, x_2 = 0, y_2 = \frac{5}{2}\)
By distance formula,
\(d(R, S) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{(0-0)^2+(\frac{5}{2}-(-3))^2}\)
\( = \sqrt{(\frac{5}{2}+3)^2}\)
\( = \sqrt{(\frac{5+6}{2})^2}\)
\( = \sqrt{(\frac{11}{2})^2}\)
\(d(R, S) = \frac{11}{2}\) units
The distance between the points R and S is \(\frac{11}{2}\) units.
(iv) Let L (\(x_1\), \(y_1\)) and M (\(x_2\), \(y_2\)) be the given points.
\(x_1 = 5, y_1 = -8, x_2 = -7, y_2 = -3\)
By distance formula,
\(d(L, M) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{(-7-5)^2 +[-3-(-8)]^2}\)
\( = \sqrt{(-7-5)^2+(-3+8)^2}\)
\( = \sqrt{(-12)^2 + (5)^2}\)
\( = \sqrt{144+25}\)
\( = \sqrt{169}\)
\(d(L, M) = 13\) units
The distance between the points L and M is 13 units.
(v) Let T (\(x_1,y_1\)) and R (\(x_2, y_2\)) be the given points.
\(x_1 = -3, y_1 = 6, x_2 = 9, y_2 = -10\)
By distance formula,
\(d(T, R) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{[9-(-3)]^2+(-10-6)^2}\)
\( = \sqrt{(9+3)^2+(-10-6)^2}\)
\( = \sqrt{12^2 + (-16)^2}\)
\( = \sqrt{144+256}\)
\( = \sqrt{400}\)
\(d(T, R) = 20\) units
The distance between the points T and R 20 units.
(vi) Let W (\(x_1\), \(y_1\)) and X (\(x_2\), \(y_2\)) be the given points.
\(x_1 = -\frac{7}{2}, y_1 = 4, x_2 = 11, y_2 = 4\)
By distance formula,
\(d(W, X) = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
\( = \sqrt{[11-(-\frac{7}{2})]^2+(4-4)^2}\)
\( = \sqrt{(11+\frac{7}{2})^2}\)
\( = \sqrt{(\frac{22+7}{2})^2}\)
\( = \sqrt{(\frac{29}{2})^2}\)
\(d(W, X) = \frac{29}{2}\) units
The distance between the points W and X is \(\frac{29}{2}\) units.
In simple words: To find the distance between two points, apply the distance formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \). Substitute the given coordinates, perform the subtractions and squaring, then add the results and take the square root to get the final distance.

🎯 Exam Tip: Ensure careful substitution of coordinates into the distance formula, especially with negative numbers and fractions, as calculation errors are common. Simplify square roots fully for final answers.

Question 2. Determine whether the points are collinear.
(i) A (1, -3), B (2, -5), C (-4, 7)
(ii) L (-2, 3), M (1, -3), N (5, 4)
(iii) R (0, 3), D (2, 1), S (3, -1)
(iv) P (-2, 3), Q (1, 2), R (4, 1)
Answer:
(i) By distance formula,
\(d(A, B) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{(2-1)^2+[-5-(-3)]^2}\)
\( = \sqrt{(2-1)^2+(-5+3)^2}\)
\( = \sqrt{1^2 + (-2)^2}\)
\( = \sqrt{1+4}\)
\(d(A, B) = \sqrt{5}\) ...(i)
\(d(B, C) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{(-4-2)^2 +[7-(-5)]^2}\)
\( = \sqrt{(-4-2)^2+(7+5)^2}\)
\( = \sqrt{(-6)^2 +12^2}\)
\( = \sqrt{36+144}\)
\( = \sqrt{180}\)
\(d(B, C) = 6\sqrt{5}\) ...(ii)
\(d(A, C) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{(-4-1)^2 +[7-(-3)]^2}\)
\( = \sqrt{(-4-1)^2+(7+3)^2}\)
\( = \sqrt{(-5)^2 +10^2}\)
\( = \sqrt{25+100}\)
\( = \sqrt{125}\)
On adding (i) and (iii),
\(d(A, B) + d(A, C)= \sqrt{5} + 5\sqrt{5} = 6\sqrt{5}\)
\(d(A, B) + d(A, C) = d(B, C)\) ... [From (ii)]
Points A, B and C are collinear.
(ii) By distance formula,
\(d(L, M) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{[1-(-2)]^2+(-3-3)^2}\)
\( = \sqrt{(1+2)^2+(-3-3)^2}\)
\( = \sqrt{3^2 +(-6)^2}\)
\( = \sqrt{9+36}\)
\(d(L, M) = \sqrt{45} = 3\sqrt{5}\) ...(i)
\(d(M, N) = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
\( = \sqrt{(5-1)^2 +[4-(-3)]^2}\)
\( = \sqrt{(5-1)^2+(4+3)^2}\)
\( = \sqrt{4^2+7^2}\)
\( = \sqrt{16+49}\)
\(d(M, N) = \sqrt{65}\) ...(ii)
\(d(L, N) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{[5-(-2)]^2+(4-3)^2}\)
\( = \sqrt{(5+2)^2+(4-3)^2}\)
\( = \sqrt{7^2 + 1^2}\)
\( = \sqrt{49+1}\)
\(d(L, N) = \sqrt{50} = 5\sqrt{2}\) ...(iii)
On adding (i) and (iii),
\(d(L, M) + d(L, N) = 3\sqrt{5} + 5\sqrt{2} \ne \sqrt{65}\)
\(d(L, M) + d(L, N) \ne d(M, N)\) ... [From (ii)]
Points L, M and N are not collinear.
(iii) By distance formula,
\(d(R, D) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{(2-0)^2+(1-3)^2}\)
\( = \sqrt{2^2+(-2)^2}\)
\( = \sqrt{4+4}\)
\(d(R, D) = \sqrt{8}\) ...(i)
\(d(D, S) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{(3-2)^2+(-1-1)^2}\)
\( = \sqrt{1^2 + (-2)^2}\)
\( = \sqrt{1+4}\)
\(d(D, S) = \sqrt{5}\) ...(ii)
\(d(R, S) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{(3-0)^2+(-1-3)^2}\)
\( = \sqrt{3^2 + (-4)^2}\)
\( = \sqrt{9+16}\)
\(d(R, S) = \sqrt{25} = 5\) ...(iii)
On adding (i) and (ii),
\(d(R, D) + d(D, S) = \sqrt{8} + \sqrt{5} \ne 5\)
\(d(R, D) + d(D, S) \ne d(R, S)\) ... [From (iii)]
Points R, D and S are not collinear.
(iv) By distance formula,
\(d(P, Q) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{[1-(-2)]^2+(2-3)^2}\)
\( = \sqrt{(1+2)^2+(2-3)^2}\)
\( = \sqrt{3^2+(-1)^2}\)
\( = \sqrt{9+1}\)
\(d(P, Q) = \sqrt{10}\) ...(i)
\(d(Q, R) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{(4-1)^2+(1-2)^2}\)
\( = \sqrt{3^2 + (-1)^2}\)
\( = \sqrt{9+1}\)
\(d(Q, R) = \sqrt{10}\) ...(ii)
\(d(P, R) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\( = \sqrt{[4-(-2)]^2+(1-3)^2}\)
\( = \sqrt{(4+2)^2+(1-3)^2}\)
\( = \sqrt{6^2+(-2)^2}\)
\( = \sqrt{36+4}\)
\( = \sqrt{40}\)
\(d(P, R) = 2\sqrt{10}\) ...(iii)
On adding (i) and (ii),
\(d(P, Q) + d(Q, R) = \sqrt{10} + \sqrt{10} = 2\sqrt{10}\)
\(d(P, Q) + d(Q, R) = d(P, R)\) ... [From (iii)]
Points P, Q and R are collinear.
In simple words: To determine if points are collinear, calculate the distances between all three pairs of points. If the sum of the two shorter distances equals the longest distance, the points are collinear; otherwise, they are not.

🎯 Exam Tip: Accurately calculate all three distances using the distance formula. A common error is misinterpreting the collinearity condition, so explicitly check if the sum of two distances equals the third.

Question 3. Find the point on the X-axis which is equidistant from A (-3,4) and B (1, -4).
Answer:
Let point C be on the X-axis which is equidistant from points A and B.
Point C lies on X-axis.
Its y co-ordinate is 0.
Let C = (x, 0)
C is equidistant from points A and B.
AC = BC
\[\sqrt{[x - (-3)]^2 + (0 - 4)^2} = \sqrt{(x - 1)^2 + [0 - (-4)]^2}\] ...[By distance formula]
\[[x-(-3)]^2 + (0-4)^2 = (x - 1)^2 + [0-(-4)]^2\] ...[Squaring both sides]
\((x + 3)^2 + (-4)^2 = (x-1)^2 + 4^2\)
\(x^2 + 6x + 9 + 16 = x^2 - 2x + 1 + 16\)
\(8x = - 8\)
\(x = \frac{-8}{8} = -1\)
The point on X-axis which is equidistant from points A and B is (-1,0).
In simple words: To find a point on the X-axis equidistant from two given points, assume the point is \((x, 0)\). Set the distance from \((x, 0)\) to the first point equal to the distance from \((x, 0)\) to the second point using the distance formula, then solve the resulting equation for \(x\).

🎯 Exam Tip: When finding a point on the X-axis, its y-coordinate is always 0. Remember to square both sides of the distance equation to eliminate the square roots, which simplifies the algebraic solution.

Question 4. Verify that points P (-2, 2), Q (2, 2) and R (2, 7) are vertices of a right angled triangle.
Answer:
Distance between two points
\( = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
By distance formula,
\(PQ = \sqrt{[2-(-2)]^2 +(2-2)^2}\)
\( = \sqrt{(2+2)^2+(0)^2} = \sqrt{(4)^2} = 4\) ...(i)
\(QR = \sqrt{(2-2)^2+(7-2)^2}\)
\( = \sqrt{(0)^2 +(5)^2} = \sqrt{(5)^2} = 5\) ...(ii)
\(PR = \sqrt{[2-(-2)]^2+(7-2)^2}\)
\( = \sqrt{(2+2)^2+(5)^2} = \sqrt{(4)^2+(5)^2}\)
\( = \sqrt{16+25} = \sqrt{41}\)
Now, \(PR^2 = (\sqrt{41})^2 = 41\) ...(iii)
Consider, \(PQ^2 + QR^2 = 4^2 + 5^2 = 16 + 25 = 41\) [From (i) and (ii)]
\(PR^2 = PQ^2 + QR^2\) ... [From (iii)]
\(\triangle PQR\) is a right angled triangle. ... [Converse of Pythagoras theorem]
Points P, Q and R are the vertices of a right angled triangle.
In simple words: To verify if points form a right-angled triangle, calculate the square of the lengths of all three sides using the distance formula. If the square of the longest side equals the sum of the squares of the other two sides (Pythagoras theorem), then it's a right-angled triangle.

🎯 Exam Tip: Remember the converse of the Pythagoras theorem: if \(a^2 + b^2 = c^2\), then the triangle is right-angled. Ensure all distances are calculated and squared correctly before applying this check.

Question 5. Show that points P (2, -2), Q (7, 3), R (11, -1) and S (6, -6) are vertices of a parallelogram.
Answer:
Proof:
Distance between two points
\( = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
By distance formula,
\(PQ = \sqrt{(7-2)^2 +[3-(-2)]^2}\)
\( = \sqrt{(7-2)^2+(3+2)^2}\)
\( = \sqrt{5^2+5^2} = \sqrt{25+25} = \sqrt{50}\) ...(i)
\(QR = \sqrt{(11-7)^2+(-1-3)^2}\)
\( = \sqrt{4^2 +(-4)^2} = \sqrt{16+16} = \sqrt{32}\) ...(ii)
\(RS = \sqrt{(6-11)^2 +[-6-(-1)]^2}\)
\( = \sqrt{(6-11)^2 + (-6 + 1)^2}\)
\( = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 +25} = \sqrt{50}\) ...(iii)
\(PS = \sqrt{(6-2)^2+[-6-(-2)]^2}\)
\( = \sqrt{(6-2)^2 + (-6+2)^2}\)
\( = \sqrt{4^2 +(-4)^2} = \sqrt{16+16} = \sqrt{32}\) ...(iv)
\(PQ = RS\) ... [From (i) and (iii)]
\(QR = PS\) ... [From (ii) and (iv)]
A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.
\( \Box PQRS\) is a parallelogram.
Points P, Q, R and S are the vertices of a parallelogram.
In simple words: To prove a quadrilateral is a parallelogram, calculate the lengths of all four sides using the distance formula. If both pairs of opposite sides are equal in length, then the figure is a parallelogram.

🎯 Exam Tip: Clearly label your distance calculations (e.g., PQ, QR, RS, PS). The key criterion for a parallelogram is that opposite sides must be congruent, so demonstrate this equality in your proof.

Question 6. Show that points A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are vertices of rhombus ABCD.
Answer:
Proof:
Distance between two points
\( = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
By distance formula,
\(AB = \sqrt{[-1-(-4)]^2+[2-(-7)]^2}\)
\( = \sqrt{(-1+4)^2 + (2+7)^2}\)
\( = \sqrt{3^2 +9^2} = \sqrt{9+81} = \sqrt{90}\) ...(i)
\(BC = \sqrt{[8-(-1)]^2+(5-2)^2}\)
\( = \sqrt{(8 + 1)^2 + (5-2)^2}\)
\( = \sqrt{9^2 + 3^2} = \sqrt{81+9} = \sqrt{90}\) ...(ii)
\(CD = \sqrt{(5-8)^2+(-4-5)^2}\)
\( = \sqrt{(-3)^2 + (-9)^2} = \sqrt{9+81} = \sqrt{90}\) ...(iii)
\(AD = \sqrt{[5-(-4)]^2+[-4-(-7)]^2}\)
\( = \sqrt{(5+4)^2+(-4+7)^2}\)
\( = \sqrt{9^2 + 3^2} = \sqrt{81+9} = \sqrt{90}\) ...(iv)
\(AB = BC = CD = AD\) ...[From (i), (ii), (iii) and (iv)]
In a quadrilateral, if all the sides are equal, then it is a rhombus.
\( \Box ABCD\) is a rhombus.
Points A, B, C and D are the vertices of rhombus ABCD.
In simple words: To prove a quadrilateral is a rhombus, calculate the lengths of all four sides using the distance formula. If all four sides are equal in length, then the figure is a rhombus.

🎯 Exam Tip: A rhombus is defined by having all four sides of equal length. Make sure your distance calculations are precise to show this equality for all sides.

Practice Set 5.1

Question 7. Find x if distance between points L (x, 7) and M (1,15) is 10.
Answer:
\(x_1 = x, y_1 = 7, x_2 = 1, y_2 = 15\)
By distance formula,
\(d(L, M) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\(d(L, M) = \sqrt{(1-x)^2 + (15-7)^2}\)
\(10 = \sqrt{(1-x)^2 +8^2}\)
\(100 = (1-x)^2+64\) ... [Squaring both sides]
\((1-x)^2 = 100-64\)
\((1-x)^2 = 36\)
\(1-x = \pm\sqrt{36}\)
... [Taking square root of both sides]
\(1-x = \pm 6\)
\(1-x = 6\) or \(1-x = -6\)
\(x = - 5\) or \(x = 7\)
The value of x is - 5 or 7.
In simple words: Given the distance between two points and one unknown coordinate, substitute the known values into the distance formula. Square both sides to remove the square root, then solve the resulting quadratic equation for the unknown variable.

🎯 Exam Tip: When squaring both sides to remove the square root, remember that the result of the square root can be positive or negative, leading to two possible solutions for the unknown variable.

Geometry 5.1

Question 8. Show that the points A (1, 2), B (1, 6), C \( (1 + 2\sqrt{3}, 4) \) are vertices of an equilateral triangle.
Answer:
Proof:
Distance between two points
\( = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
By distance formula,
\(AB = \sqrt{(1-1)^2 +(6-2)^2} = \sqrt{0^2 + 4^2}\)
\( = \sqrt{4^2} = 4\) ...(i)
\(BC = \sqrt{(1+2\sqrt{3}-1)^2+(4-6)^2}\)
\( = \sqrt{(2\sqrt{3})^2+(-2)^2}\)
\( = \sqrt{12+4} = \sqrt{16} = 4\) ...(ii)
\(AC = \sqrt{(1+2\sqrt{3}-1)^2+(4-2)^2}\)
\( = \sqrt{(2\sqrt{3})^2+2^2}\)
\( = \sqrt{12+4} = \sqrt{16} = 4\) ...(iii)
\(AB = BC = AC\) ... [From (i), (ii) and (iii)]
\(\triangle ABC\) is an equilateral triangle.
Points A, B and C are the vertices of an equilateral triangle.
In simple words: To prove that three points form an equilateral triangle, calculate the lengths of all three sides using the distance formula. If all three side lengths are equal, then the triangle is equilateral.

🎯 Exam Tip: An equilateral triangle has three equal sides. Carefully calculate each side's length, especially when coordinates involve square roots, to ensure accuracy in demonstrating equality.

Question 1. In the figure, seg AB || Y-axis and seg CB || X-axis. Co-ordinates of points A and C are given. To find AC, fill in the boxes given below. (Textbook pa. no. 102)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक कार्टेशियन समन्वय प्रणाली दिखाता है जिसमें X और Y अक्ष हैं। इसमें तीन बिंदु A (2, 3), B (2, 2) और C (-2, 2) अंकित हैं। खंड AB Y-अक्ष के समानांतर है और खंड CB X-अक्ष के समानांतर है, जो दर्शाता है कि त्रिभुज ABC बिंदु B पर समकोण है।
Answer:
Solution:
In \(\triangle ABC, \angle B = 900\)
\((AB)^2 + (BC)^2 = [(AC)^2]\) ...(i) ... [Pythagoras theorem]
seg CB || X-axis
y co-ordinate of B = 2
seg BA || Y-axis
x co-ordinate of B = 2
co-ordinate of B is (2, 2) = (\(x_1,y_1\))
co-ordinate of A is (2, 3) = (\(x_2,y_2\))
Since, AB || to Y-axis,
d(A, B) = \(Y_2 - Y_1\)
d(A,B) = \(3 - 2 = 1\)
co-ordinate of C is (-2,2) = (\(x_1,y_1\))
co-ordinate of B is (2, 2) = (\(x_2,y_2\))
Since, BC || to X-axis,
d(B, C) = \(X_2 - X_1\)
d(B,C) = \(2 - (-2) = 4\)
\(AC^2 = 1^2 + 4^2\) ... [From (i)]
\( = 1 + 16 = 17\)
\(AC = \sqrt{17}\) units ... [Taking square root of both sides]
In simple words: For a right-angled triangle on a coordinate plane, calculate the lengths of the legs parallel to the axes by finding the difference in coordinates. Then, use the Pythagorean theorem to find the length of the hypotenuse.

🎯 Exam Tip: When sides are parallel to axes, distance calculation simplifies to absolute differences in coordinates. Ensure correct application of the Pythagorean theorem: \((\text{hypotenuse})^2 = (\text{leg}_1)^2 + (\text{leg}_2)^2\).