CBSE Class 10 Mathematics Coordinate Geometry Worksheet Set 04

Question. The distance between the points \( A(p \sin 25^{\circ}, 0) \) and \( B(0, p \sin 65^{\circ}) \) is
(a) 0 units
(b) p units
(c) \( p^2 \) units
(d) 1 units
Answer: (b) p units
Explanation: The distance between point A and point B=
\( AB = \sqrt{(0 - p \sin 25^{\circ})^2 + (p \sin 65^{\circ} - 0)^2} \)
\( = \sqrt{p^2 \sin^2 25^{\circ} + p^2 \sin^2 65^{\circ}} \)
\( = p \sqrt{\sin^2 25^{\circ} + \sin^2 (90^{\circ} - 25^{\circ})} \)
\( = p \sqrt{\sin^2 25^{\circ} + \cos^2 25^{\circ}} \) [\( \because \sin(90^{\circ} - \theta) = \cos \theta \)]
= p units
[\( \because \cos^2 \theta + \sin^2 \theta = 1 \)]

 

Question. If the points (x, y), (1, 2) and (7, 0) are collinear, then the relation between ‘x’ and ‘y’ is given by
(a) 3x – y – 7 = 0
(b) 3x + y + 7 = 0
(c) x + 3y – 7 = 0
(d) x – 3y + 7 = 0
Answer: (c) x + 3y – 7 = 0
Explanation: \( \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0 \)
\( \implies \frac{1}{2} |x(2 - 0) + 1(0 - y) + 7(y - 2)| = 0 \)
\( \implies \frac{1}{2} |2x - y + 7y - 14| = 0 \)
\( \implies 2x + 6y - 14 = 0 \)
\( \implies x + 3y - 7 = 0 \)

 

Question. If the distance between the points (p, – 5) and (2, 7) is 13 units, then the value of ‘p’ is
(a) -3, -7
(b) 3, -7
(c) 3, 7
(d) -3, 7
Answer: (d) -3, 7
Explanation: Let point A be (p, -5) and point B (2, 7) and distance between A and B = 13 units
\( \therefore 13 = \sqrt{(2 - p)^2 + (7 + 5)^2} \)
\( \implies 13 = \sqrt{4 + p^2 - 4p + 144} \)
\( \implies 13 = \sqrt{p^2 - 4p + 148} \)
\( \implies 169 = p^2 - 4p + 148 \)
\( \implies p^2 - 4p - 21 = 0 \)
\( = p^2 - 7p + 3p - 21 = 0 \)
\( = p(p - 7) + 3(p - 7) = 0 \)
\( \implies (p - 7)(p + 3) = 0 \)
\( \implies p = 7, p = -3 \)

 

Question. If the vertices of a triangle are (1, 1), ( – 2, 7) and (3, – 3), then its area is
(a) 0 sq. units
(b) 2 sq. units
(c) 24 sq. units
(d) 12 sq. units
Answer: (a) 0 sq. units
Explanation: Given: \( (x_1, y_1) = (1, 1) \), \( (x_2, y_2) = (-2, 7) \) and \( (x_3, y_3) = (3, -3) \), then the Area of triangle
\( = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
\( = \frac{1}{2} |1(7 + 3) + (-2)(-3 - 1) + 3(1 - 7)| \)
\( = \frac{1}{2} |10 + 8 - 18| \)
\( = \frac{1}{2} |0| = 0 \) sq. units
Also therefore the three given points(vertices) are collinear.

 

Question. The distance between the points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by
(a) \( \sqrt{(x_2 + x_1)^2 + (y_2 + y_1)^2} \) units
(b) \( \sqrt{(x_2 + x_1)^2 - (y_2 + y_1)^2} \) units
(c) \( \sqrt{(x_2 - x_1)^2 - (y_2 - y_1)^2} \) units
(d) \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) units
Answer: (d) \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) units
Explanation: The distance between the points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) units. This is known as distance formula.

 

Question. If the points A(x, 2), B(- 3, - 4), C(7, - 5) are collinear, then find the value of x.
Answer: Since the points are collinear, then,
Area of triangle = 0
\( \frac{1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] = 0 \)
\( \frac{1}{2} [x(-4 + 5) + (-3)(-5 - 2) + 7(2 + 4)] = 0 \)
x + 21 + 42 = 0
x = -63

 

Question. Find the distance between the points A and B in the following : A(a, 0), B(0, a)
Answer: A(a, 0), B(0, a)
\( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(0 - a)^2 + (a - 0)^2} \)
\( = \sqrt{a^2 + a^2} = \sqrt{2a^2} = \sqrt{2}a \) units

 

Question. Find the perpendicular distance of A(5,12) from the y-axis.
Answer: The point on the y-axis is (0,12)
\( \therefore \) Distance between (5,12) and (0,12)
\( d = \sqrt{(0 - 5)^2 + (12 - 12)^2} \)
\( = \sqrt{25 + 0} \)
= 5 units

 

Question. Find the distance of the point (- 4, - 7) from the y-axis.
Answer: Points are (- 4, - 7) and (0, - 7)
Distance \( = \sqrt{(0 + 4)^2 + (-7 + 7)^2} \)
\( = \sqrt{4^2 + 0} = \sqrt{16} = 4 \) units

 

Question. Find the coordinates of the centroid of a triangle whose vertices are (0,6), (8,12) and (8,0).
Answer: Coordinates of the centroid of a triangle whose vertices are \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) are
\( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)
\( = \left( \frac{0 + 8 + 8}{3}, \frac{6 + 12 + 0}{3} \right) = \left( \frac{16}{3}, \frac{18}{3} \right) = \left( \frac{16}{3}, 6 \right) \).

 

Question. Find the distance between the points: A(-6, -4) and B(9, -12)
Answer: The given points are A(-6, -4) and B(9, -12)
Then, \( (x_1 = -6, y_1 = -4) \) and \( (x_2 = 9, y_2 = -12) \)
\( \therefore AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( = \sqrt{(9 + 6)^2 + (-12 + 4)^2} = \sqrt{(15)^2 + (-8)^2} \)
\( = \sqrt{225 + 64} = \sqrt{289} = 17 \) units

 

Question. Find the condition that the point (x, y) may lie on the line joining (3, 4) and (-5, - 6).
Answer: Since the point P (x, y) lies on the line joining A (3, 4) and B (-5, -6). Therefore, P, A and B are collinear points.
\( \therefore \{4x + 3 \times -6 + (-5) \times y\} - \{3y + (-5) \times 4 + x \times (-6)\} = 0 \)
\( \implies \{4x - 18 - 5y) - (3y - 20 - 6x) = 0 \)
\( \implies 10x - 8y + 2 = 0 \)
\( \implies 5x - 4y + 1 = 0 \)
Hence, the point (x, y) lies on the line joining (3,4) and (-5, -6), if 5x - 4y + 1 = 0

 

Question. If P (x, y) is any point on the line joining the points A(a,0) and B(0, b), then show that \( \frac{x}{a} + \frac{y}{b} = 1 \).
Answer: It is given that the point P (x, y) lies on the line segment joining points A (a, 0) and B (0, b).
Therefore, points P (x, y), A (a, 0) and B (0, b) are collinear points.
\( \therefore (x \times 0 + a \times b + 0 \times y) - (a \times y + 0 \times 0 + x \times b) = 0 \)
\( \implies ab - (ay + bx) = 0 \)
\( \implies ab = ay + bx \)
\( \implies \frac{ab}{ab} = \frac{ay}{ab} + \frac{bx}{ab} \) [Dividing throughout by ab]
\( \implies 1 = \frac{y}{b} + \frac{x}{a} \) or \( \frac{x}{a} + \frac{y}{b} = 1 \)

 

Question. The area of triangle formed by the points (p, 2 - 2p), (1 - p, 2p) and (-4 - p, 6 - 2p) is 70 sq. units. How many integral values of p are possible.
Answer: Area \( = \frac{1}{2} [p(2p - (6 - 2p)) + (1 - p)((6 - 2p) - (2 - 2p)) + (-4 - p)((2 - 2p) - 2p)] = 70 \)
\( \implies \frac{1}{2} [p(4p - 6) + (1 - p)4 + (-4 - p)(2 - 4p)] = 70 \)
\( \implies 4p^2 - 6p + 4 - 4p - 8 + 16p - 2p + 4p^2 = 140 \)
\( \implies \frac{1}{2} [-13k - 9] = 15 \)
\( \implies [-13k - 9] = 30 \implies -13k - 9 = 30 \) or \( -13k - 9 = -30 \)
k = - 3 or k = \( \frac{21}{13} \)
When k = - 3, coordinates = 15 sq. units
\( \implies \frac{1}{2} \times AB \times Altitude = 15 \)
\( \implies \frac{1}{2} \times 3 \times Altitude = 15 \)
\( \implies Altitude = 10 \) units

 

Question. Point A is on x-axis, point B is on y-axis and the point P lies on line segment AB, such that P (4, - 5) and AP : PB = 5 : 3. Find the coordinates of point A and B.
Answer: Let coordinates of A are (x, 0) and coordinates of B are (0, y)
Using section formula, we get
\( 4 = \frac{5 \times 0 + 3 \times x}{5 + 3} \)
\( \implies 32 = 3x \)
\( \implies x = \frac{32}{3} \)
Similarly, \( 5 = \frac{5 \times y + 3 \times 0}{5 + 3} \)
\( \implies 40 = 5y \)
\( \implies y = 8 \)
\( \therefore \) Coordinate of A are \( (\frac{32}{3}, 0) \) and coordinates of B are (0, 8).

 

Question. Show that four points (0, -1), (6, 7), (-2, 3) and (8, 3) are the vertices of a rectangle. Also, find its area.
Answer: Let A (0, -1), B (6, 7), C (-2, 3) and D (8, 3) be the given points. Then,
\( AD = \sqrt{(8 - 0)^2 + (3 + 1)^2} = \sqrt{64 + 16} = 4\sqrt{5} \)
\( BC = \sqrt{(6 + 2)^2 + (7 - 3)^2} = \sqrt{64 + 16} = 4\sqrt{5} \)
\( AC = \sqrt{(-2 - 0)^2 + (3 + 1)^2} = \sqrt{4 + 16} = 2\sqrt{5} \)
and, \( BD = \sqrt{(8 - 6)^2 + (3 - 7)^2} = \sqrt{4 + 16} = 2\sqrt{5} \)
Therefore, AD = BC and AC = BD
So, ADBC is a parallelogram
Now, \( AB = \sqrt{(6 - 0)^2 + (7 + 1)^2} = \sqrt{36 + 64} = 10 \)
and, \( CD = \sqrt{(8 + 2)^2 + (3 - 3)^2} = 10 \)
Clearly, \( AB^2 = AD^2 + DB^2 \) and \( CD^2 = CB^2 + BD^2 \)
Hence, ADBC is a rectangle.
Area of rectangle ADBC = \( AD \times DB = (4\sqrt{5} \times 2\sqrt{5}) \) sq. units = 40 sq. units.

 

Question. Find the co-ordinates of the points of trisection of the line segment joining the points A(1, - 2) and B(- 3,4).
Answer: Let \( P(x_1, y_1), Q(x_2, y_2) \) divides AB into 3 equal parts.
\( \therefore \) P divides AB in the ratio of 1: 2
\( \therefore x_1 = \frac{1 \times -3 + 2 \times 1}{1 + 2} \) and \( y_1 = \frac{1 \times 4 + 2 \times -2}{1 + 2} \)
\( \implies x_1 = \frac{-3 + 2}{3} = \frac{-1}{3} \)
\( y_1 = \frac{-4 + 4}{3} = 0 \)
\( \therefore \) Co-ordinates of P \( (-\frac{1}{3}, 0) \).
Q is the mid-point of PB.
\( \therefore x_2 = \frac{\frac{-1}{3} + (-3)}{2} \)
\( = \frac{\frac{-10}{3}}{2} = \frac{-5}{3} \)
\( y_2 = \frac{0 + 4}{2} = 2 \)
\( \therefore \) Co-ordinates of Q \( (-\frac{5}{3}, 2) \).

 

Question. Show that the points A(3, 5), B(6, 0), C(1, -3) and D (-2, 2) are the vertices of a square ABCD.
Answer: Let A(3,5), B(6, 0), C(1, -3) and D(-2, 2) be the angular points of a quadrilateral ABCD.
Join AC and BD
Now \( AB = \sqrt{(6 - 3)^2 + (0 - 5)^2} \)
\( = \sqrt{3^2 + (-5)^2} \)
\( = \sqrt{9 + 25} = \sqrt{34} \) units,
\( BC = \sqrt{(1 - 6)^2 + (-3 - 0)^2} = \sqrt{(-5)^2 + (-3)^2} \)
\( = \sqrt{25 + 9} = \sqrt{34} \) units,
\( CD = \sqrt{(-2 - 1)^2 + (2 + 3)^2} = \sqrt{(-3)^2 + 5^2} \)
\( = \sqrt{9 + 25} = \sqrt{34} \) units,
and \( DA = \sqrt{(3 + 2)^2 + (5 - 2)^2} = \sqrt{5^2 + 3^2} \)
\( = \sqrt{25 + 9} = \sqrt{34} \) units,
Thus, AB = BC = CD = DA.
Diagonal \( AC = \sqrt{(1 - 3)^2 + (-3 - 5)^2} = \sqrt{(-2)^2 + (-8)^2} \)
\( = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17} \) units
Diagonal \( BD = \sqrt{(-2 - 6)^2 + (2 - 0)^2} \)
\( = \sqrt{(-8)^2 + 2^2} = \sqrt{64 + 4} \)
\( = \sqrt{68} = 2\sqrt{17} \) units
\( \therefore \) diag. AC = diag. BD
Thus, ABCD is a quadrilateral in which all sides are equal and the diagonals are equal.
Hence, quad. ABCD is a square.

 

Question. A (4, 2), B (6, 5) and C (1, 4) are the vertices of \( \triangle \) ABC.
i. The median from A meets BC in D. Find the coordinates of the point D.
ii. Find the coordinates of point P on AD such that AP : PD = 2:1.
iii. Find the coordinates of the points Q and R on medians BE and CP respectively such that BQ : QE = 2 :1 and CR: RF =2: 1.
iv. What do you observe?
Answer: i. Median AD of the triangle will divide the side BC in two equal parts. So D is the midpoint of side BC.
Coordinates of \( D = \left( \frac{6 + 1}{2}, \frac{5 + 4}{2} \right) = \left( \frac{7}{2}, \frac{9}{2} \right) \)
ii. Point P divides the side AD in a ratio 2 : 1.
Coordinates of \( P = \left( \frac{2 \times \frac{7}{2} + 1 \times 4}{2 + 1}, \frac{2 \times \frac{9}{2} + 1 \times 2}{2 + 1} \right) \)
\( = \left( \frac{11}{3}, \frac{11}{3} \right) \)
iii. Median BE of the triangle will divide the side AC in two equal parts. So E is the midpoint of side AC.
Coordinates of \( E = \left( \frac{4 + 1}{2}, \frac{2 + 4}{2} \right) = \left( \frac{5}{2}, 3 \right) \)
Point Q divides the side BE in a ratio 2:1
Coordinates of \( Q = \left( \frac{2 \times \frac{5}{2} + 1 \times 6}{2 + 1}, \frac{2 \times 3 + 1 \times 5}{2 + 1} \right) = \left( \frac{11}{3}, \frac{11}{3} \right) \)
Median CF of the triangle will divide the side AB in two equal parts. So F is the midpoint of side AB.
Coordinates of \( F = \left( \frac{4 + 6}{2}, \frac{2 + 5}{2} \right) = \left( 5, \frac{7}{2} \right) \)
Point R divides the side CF in a ratio 2:1.
Coordinates of \( R = \left( \frac{2 \times 5 + 1 \times 1}{2 + 1}, \frac{2 \times \frac{7}{2} + 1 \times 4}{2 + 1} \right) = \left( \frac{11}{3}, \frac{11}{3} \right) \)
iv. Now we may observe that coordinates of point P, Q are same. So, all these are representing same point on the plane i.e. centroid of the triangle.

 

Question. Find the lengths of the medians of a \( \triangle \) ABC whose vertices are A(0, -1) B(2, 1) and C(0, 3).
Answer: Let D, E, F be the midpoint of the side BC, CA and AB respectively in \( \triangle \) ABC
Then, by the midpoint formula, we have
\( D \left( \frac{2 + 0}{2}, \frac{1 + 3}{2} \right), E \left( \frac{0 + 0}{2}, \frac{3 - 1}{2} \right), F \left( \frac{0 + 2}{2}, \frac{-1 + 1}{2} \right) \)
i.e., D(1, 2), E(0, 1), F(1, 0)
Hence the lengths of medians AD, BE and CF are given by
\( AD = \sqrt{(1 - 0)^2 + (2 + 1)^2} = \sqrt{1 + 9} = \sqrt{10} \) units
\( BE = \sqrt{(0 - 2)^2 + (1 - 1)^2} = \sqrt{4 + 0} = \sqrt{4} = 2 \) units
\( CF = \sqrt{(1 - 0)^2 + (0 - 3)^2} = \sqrt{1 + 9} = \sqrt{10} \) units
Hence, \( AD = \sqrt{10}, BE = 2, CF = \sqrt{10} \)

 

Question. If one end of a diameter of a circle is (4, 6) and the centre is ( – 4, 7), then the other end is
(a) ( – 12, 8)
(b) (8, – 12)
(c) (8, 10)
(d) (8, – 6)
Answer: (a) ( – 12, 8)
Explanation: one end of a diameter is \( A(4, 6) \) and the centre is \( O(- 4, 7) \). .... (Given)
Let the other end be B
therefore coordinates of centre O are \( x = \frac{(4+x)}{2} \)
\( \therefore -4 = \frac{4+x}{2} \)

\( \implies 4 + x = -8 \)
\( \implies x = -12 \)
And \( y = \frac{6+y}{2} \)
7 = (6 + y) / 2

\( \implies 6 + y = 14 \)
\( \implies y = 8 \)
Therefore, the required coordinates of other ends of the diameter are \( (-12, 8) \).

 

Question. The point where the perpendicular bisector of the line segment joining the points A(2, 5) and B(4, 7) cuts is:
(a) (3, 6)
(b) (0, 0)
(c) (2, 5)
(d) (6, 3)
Answer: (a) (3, 6)
Explanation: Since, the point, where the perpendicular bisector of a line segment joining the points \( A(2, 5) \) and \( B(4, 7) \) cuts, is the mid-point of that line segment.
\( \therefore \) Coordinates of Mid-point of line segment \( AB = \left( \frac{2+4}{2}, \frac{5+7}{2} \right) = (3, 6) \)

 

Question. The point ( – 3, 5) lies in the ___________ quadrant
(a) IV
(b) II
(c) III
(d) I
Answer: (b) II
Explanation: Since \( x \)-coordinate is negative and \( y \)-coordinate is positive. Therefore, the point \( (- 3, 5) \) lies in II quadrant.

 

Question. If the mid – point of the line segment joining the points (a, b – 2) and ( – 2, 4) is (2, – 3), then the values of ‘a’ and ‘b’ are
(a) 6, 8
(b) 6, – 8
(c) 4, – 5
(d) – 6, 8
Answer: (b) 6, – 8
Explanation: Let the coordinates of midpoint \( O(2, -3) \) is equidistance from the points \( A(a, b - 2) \) and \( B(-2, 4) \).
\( \therefore 2 = \frac{a-2}{2} \)

\( \implies a - 2 = 4 \)
\( \implies a = 6 \)
Also \( -3 = \frac{b-2+4}{2} \)

\( \implies b + 2 = -6 \)
\( \implies b = -8 \)
Therefore, \( a = 6 \) and \( b = -8 \).

 

Question. Find the value of ‘k’, if the point (0, 2) is equidistant from the points (3, k) and (k, 5)
(a) 2
(b) 0
(c) 1
(d) -1
Answer: (c) 1
Explanation: Let point \( C(0, 2) \) is equidistant from the points \( A(3, k) \) and \( B(k, 5) \).
i.e. AC = BC
\( \therefore AC^2 = BC^2 \)

\( \implies (3 - 0)^2 + (k - 2)^2 = (k - 0)^2 + (5 - 2)^2 \)
\( \implies 9 + k^2 + 4 - 4k = k^2 + 9 \)
\( \implies 4k = 4 \)
\( \implies k = 1 \)

 

Question. If origin is the mid-point of the line segment joined by the points (2, 3) and (x, y) then find the value of (x, y).
Answer:
\( \frac{2+x}{2} = 0 \)

\( \implies x = - 2 \)
\( \frac{3+y}{2} = 0 \)

\( \implies y = - 3 \).

 

Question. Find the number of points on x-axis which are at a distance of 2 units from (2, 4).
Answer: Distance of the point (2, 4) from x-axis is 4 units. There is no point on x-axis which is at a distance of 2 units from the given point.

 

Question. Find the perimeter of a triangle with vertices (0, 4), (0,0) and (3,0).
Answer: Here, \( A(0,4), B(0,0), C(3,0) \)
\( AB = \sqrt{(0 - 0)^2 + (0 - 4)^2} = \sqrt{16} = 4 \)
\( BC = \sqrt{(3 - 0)^2 + (0 - 0)^2} = \sqrt{9} = 3 \)
\( CA = \sqrt{(0 - 3)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)
Therefore, Perimeter of triangle = 4 + 3 + 5 = 12

 

Question. Find the distance between the points A and B in the following:A(1,-3), B(4, 1)
Answer: \( A(1, -3), B(4, 1) \)
\( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4 - 1)^2 + [1 - (-3)]^2} \)
\( = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) units

 

Question. Find the coordinates of the point , where the line x - y = 5 cuts Y-axis.
Answer: \( x - y = 5 \) is a given line
\( x - y = 5 \) cuts Y-axis.
Put \( x = 0 \) in the equation of line \( x - y = 5 \)

\( \implies (0) - y = 5 \)
\( \implies y = -5 \)
Therefore, the point is (0,-5) cuts \( x - y = 5 \) at Y-axis.

 

Question. Find the value of 'k' if the points (7, –2), (5, 1), (3, k) are collinear.
Answer: (7, –2), (5, 1), (3, k)
Area of the triangle
\( = \frac{1}{2} [7(1 - k) + 5(k - (-2)) + 3(-2 - 1)] \)
\( = \frac{1}{2} [7 - 7k + 5k + 10 - 9] \)
\( = \frac{1}{2} [8 - 2k] = 4 - k \)
If the points are collinear, then area of the triangle = 0

\( \implies 4 - k = 0 \)
\( \implies k = 4 \)

 

Question. The point R divides the line segment AB where A(-4, 0), B(0, 6) are such that \( AR = \frac{3}{4} AB \). Find the coordinates of R.
Answer: Let coordinates of R be (x, y)
\( AR = \frac{3}{4} AB \) [Given]
But AR + RB = AB

\( \implies \frac{3}{4} AB + RB = AB \)
\( \implies RB = AB - \frac{3}{4} AB = \frac{4AB - 3AB}{4} = \frac{AB}{4} \)
\( \frac{AR}{RB} = \frac{\frac{3}{4} AB}{\frac{1}{4} AB} = \frac{3}{4} : \frac{1}{4} = \frac{3}{4} \times \frac{4}{1} = 3 : 1 \)
Thus, R divides AB in the ratio 3 : 1.
\( x = \frac{3 \times 0 + 1 \times (-4)}{3 + 1} = \frac{0 - 4}{4} = \frac{-4}{4} = - 1 \)
and \( y = \frac{3 \times 6 + 1 \times 0}{3 + 1} = \frac{18 + 0}{4} = \frac{18}{4} = \frac{9}{2} \)
Thus, coordinates of R are \( \left( -1, \frac{9}{2} \right) \)

 

Question. Find the centroid of the triangle whose vertices are given below: (3, -5), (-7, 4), (10, -2).
Answer: The given vertices of triangle are (3, -5), (-7, 4) and (10, -2).
Let (x, y) be the coordinates of the centroid. Then
\( x = \frac{x_1 + x_2 + x_3}{3} = \frac{3 + (-7) + 10}{3} \)
\( = \frac{13 - 7}{3} = \frac{6}{3} = 2 \)
\( y = \frac{y_1 + y_2 + y_3}{3} = \frac{-5 + 4 + (-2)}{3} \)
\( = \frac{-7 + 4}{3} = \frac{-3}{3} = -1 \)
\( \therefore \) The coordinates of the centroid are (2, –1)

 

Question. Prove that the lines joining the middle points of the opposite sides of a quadrilateral and the join of the middle points of its diagonals meet in a point and bisect one another.
Answer: Let OBCD be the quadrilateral P, Q, R, S be the mid-points of OB, CD, OD and BC.
Let the coordinates of O, B, C, D are (0, 0), (x, 0), (x, y) and (0, y)
Coordinates of P are \( \left( \frac{x}{2}, 0 \right) \)
Coordinates of Q are \( \left( \frac{x}{2}, y \right) \)
Coordinates of R are \( \left( 0, \frac{y}{2} \right) \)
Coordinates of S are \( \left( x, \frac{y}{2} \right) \)
Coordinates of mid-point of PQ are \( \left( \frac{\frac{x}{2} + \frac{x}{2}}{2}, \frac{0 + y}{2} \right) = \left( \frac{x}{2}, \frac{y}{2} \right) \)
Coordinates of mid-point of RS are \( \left( \frac{(0 + x)}{2}, \frac{\frac{y}{2} + \frac{y}{2}}{2} \right) = \left( \frac{x}{2}, \frac{y}{2} \right) \)
Since, the coordinates of the mid-point of PQ = coordinates of mid-point of RS.
\( \therefore \) PQ and RS bisect each other.

 

Question. Find the value of m for which the points with coordinates (3, 5), (m, 6) and \( (\frac{1}{2}, \frac{15}{2}) \) are collinear.
Answer: If points are collinear, then one point divides the other two in the same ratio.
Let point (m, 6) divides the join of (3, 5) and \( \left( \frac{1}{2}, \frac{15}{2} \right) \) in the ratio k: 1.
Then, \( (m, 6) = \left( \frac{\frac{k}{2} + 3}{k + 1}, \frac{\frac{15k}{2} + 5}{k + 1} \right) \)

\( \implies m = \frac{\frac{k}{2} + 3}{k + 1} \) ...(i)
and \( 6 = \frac{\frac{15k}{2} + 5}{k + 1} \) ...(ii)
From (ii), we get \( 6k + 6 = \frac{15k}{2} + 5 \)

\( \implies 6k - \frac{15k}{2} = - 1 \)
\( \implies -\frac{3}{2} k = - 1 \)
\( \implies k = \frac{2}{3} \)
Substituting, \( k = \frac{2}{3} \) in (i), we get
m = \( \frac{\frac{1}{2} \times \frac{2}{3} + 3}{\frac{2}{3} + 1} = \frac{\frac{1}{3} + 3}{\frac{5}{3}} = \frac{\frac{10}{3}}{\frac{5}{3}} = 2 \)
Hence, for m = 2 points are collinear.

 

Question. If the points A (a, -11), B (5, b), C (2, 15) and D (1, 1) are the vertices of a parallelogram ABCD, find the values of a and b.
Answer: Let A(a, -11), B(5, b), C(2, 15) and D(1, 1) be the given points.
We know that diagonals of parallelogram bisect each other.
Therefore, Coordinates of mid-point of AC = Coordinates of mid-point of BD
\( \left( \frac{a + 2}{2}, \frac{15 - 11}{2} \right) = \left( \frac{5 + 1}{2}, \frac{b + 1}{2} \right) \)

\( \implies \frac{a + 2}{2} = 3 \) and \( \frac{4}{2} = \frac{b + 1}{2} \)

\( \implies a + 2 = 6 \) and \( b + 1 = 4 \)

\( \implies a = 6 - 2 \) and \( b = 4 - 1 \)

\( \implies a = 4 \) and \( b = 3 \)
Hence value of a and b is equal to 4 and 3 respectively.

 

Question. In the given triangle ABC as shown in diagram D, E and F are the mid-points of AB, BC and AC respectively. Find the area of \( \triangle \) DEF.
Answer: Let \( D(x_1, y_1) \) be the mid-point of AB,then,
\( x_1 = \frac{3 - 5}{2} = -1 \) and \( y_1 = \frac{2 - 6}{2} = -2 \)
\( \therefore D = (-1, -2) \)
Let \( E(x_2, y_2) \) be the mid-point of BC,then,
\( x_2 = \frac{-5 + 7}{2} = 1 \)
and \( y_2 = \frac{-6 + 4}{2} = -1 \)
\( \therefore E = (1, -1) \)
Let \( F(x_3, y_3) \) be the mid-point of AC,then
\( x_3 = \frac{7 + 3}{2} = 5 \) and \( y_3 = \frac{4 + 2}{2} = 3 \)
Now, area \( \triangle \) DEF
\( = \frac{1}{2} [-1(-1 - 3) + 1(3 + 2) + 5(-2 + 1)] \)
\( = \frac{1}{2} [4 + 5 - 5] \)
= 2 units

 

Question. Find the area of a quadrilateral PQRS whose vertices area P(- 5, 7), Q(- 4, - 5), R (-1, - 6) and S(4, 5).
Answer: Area \( \square \) PQRS = ar \( \triangle \) PQS + ar \( \triangle \) QRS
Ar \( \triangle \) PQS \( = \frac{1}{2} [(-5)(- 5 - 5 ) + (-4)(5 - 7 ) + 4(7 + 5)] \)
\( = \frac{1}{2} [50 + 8 + 48] \)
\( = \frac{1}{2} \times 106 = 53 \) units
Ar \( \triangle \) QRS \( = \frac{1}{2} [(-4)(- 6 - 5) + (-1)(5 + 5) + 4(-5 + 6)] \)
\( = \frac{1}{2} [44 + (-10) + 4] \)
\( = \frac{1}{2} \times 38 = 19 \) units
Hence, area \( \square \) PQRS = 53 + 19 = 72 sq. units

 

Question. If the point A(2, –4) is equidistant from P(3, 8) and Q(–10, y) then find the values of y. Also find distance PQ.
Answer: According to the question,we are given that,
PA = QA

\( \implies PA^2 = QA^2 \)

\( \implies (3 – 2)^2 + (8 + 4)^2 = (–10 – 2)^2 + (y + 4)^2 \)

\( \implies 1^2 + 12^2 = (–12)^2 + y^2 + 16 + 8y \)

\( \implies y^2 + 8y + 16 – 1 = 0 \)

\( \implies y^2 + 8y + 15 = 0 \)

\( \implies y^2 + 5y + 3y + 15 = 0 \)

\( \implies y(y + 5) + 3(y + 5) = 0 \)

\( \implies (y + 5) (y + 3) = 0 \)

\( \implies y + 5 = 0 \) or \( y + 3 = 0 \)

\( \implies y = –5 \) or \( y = –3 \)
So, the co–ordinates are \( P(3, 8), Q_1(–10, –3), Q_2(–10, –5) \).
Now, \( PQ_1^2 = (3 + 10)^2 + (8 + 3)^2 = 13^2 + 11^2 \)

\( \implies PQ_1^2 = 169 + 121 \)

\( \implies PQ_1 = \sqrt{290} \) units
and \( PQ_2^2 = (3 + 10)^2 + (8 + 5)^2 = 13^2 + 13^2 = 13^2 [1 + 1] \)

\( \implies PQ_2^2 = 13^2 \times 2 \)

\( \implies PQ_2 = 13\sqrt{2} \) units
Hence, \( y = –3, –5 \) and \( PQ = \sqrt{290} \) units and \( 13\sqrt{2} \) units.

 

Question. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (-1, -2), (1, 0), (-1, 2), (-3, 0)
Answer: (-1, -2), (1, 0), (-1, 2), (-3, 0)
Let \( A \to (-1, -2), B \to (1, 0), C \to (-1, 2) \) and \( D \to (-3, 0) \)
Then, \( AB = \sqrt{[1 - (-1)]^2 + [0 - (-2)]^2} \)
\( = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \)
\( BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} \)
\( = \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \)
\( CD = \sqrt{[(-3) - (-1)]^2 + (0 - 2)^2} \)
\( = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \)
\( DA = \sqrt{[(-1) - (-3)]^2 + (-2 - 0)^2} \)
\( = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \)
\( AC = \sqrt{[(-1) - (-1)]^2 + [(2) - (-2)]^2} = 4 \)
\( BD = \sqrt{[(-3) - (1)]^2 + (0 - 0)^2} = 4 \)
Since \( AB = BC = CD = DA \) (i.e., all the four sides of the quadrilateral ABCD are equal) and \( AC = BD \) (i.e. diagonals of the quadrilateral ABCD are equal). Therefore, ABCD is a square.

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