CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 12

Question. The point on the x-axis which is equidistant from (– 4, 0) and (10, 0) is:
(a) (7, 0)
(b) (5, 0)
(c) (0, 0)
(d) (3, 0)
Answer: (d) (3, 0)
Explanation: Since, both the given points are on the x-axis, the mid-point \( \left( \frac{-4 + 10}{2}, \frac{0 + 0}{2} \right) \), i.e., (3, 0) lies on x-axis and is equidistant from (–4, 0) and (10, 0)

Question. The centre of a circle whose end points of a diameter are (– 6, 3) and (6, 4) are:
(a) (8, – 1)
(b) (4, 7)
(c) \( \left(0, \frac{7}{2}\right) \)
(d) \( \left(4, \frac{7}{2}\right) \)
Answer: (c) \( \left(0, \frac{7}{2}\right) \)

Question. The distance between the points (m, – n) and (– m, n) is:
(a) \( \sqrt{m^2 + n^2} \)
(b) m + n
(c) \( 2\sqrt{m^2 + n^2} \)
(d) \( \sqrt{2m^2 + 2n^2} \)
Answer: (c) \( 2\sqrt{m^2 + n^2} \)
Explanation: Distance between (m, –n) and (–m, n) is
= \( \sqrt{(-m - m)^2 + (n - (-n))^2} \)
(By distance formula)
= \( \sqrt{(-2m)^2 + (2n)^2} \)
= \( \sqrt{4m^2 + 4n^2} \)
= \( 2\sqrt{m^2 + n^2} \)

Question. The point which divides the line segment joining the points (7, –6) and (3, 4) in the ratio 1:2 internally, lies in the:
(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant
Answer: (d) IV quadrant
Explanation: We know that if P (x, y) divides the line segment joining \( A (x_1, y_1) \) and \( B (x_2, y_2) \) internally in the ratio m:n, then
\( x = \frac{mx_2 + nx_1}{m + n} \) and \( y = \frac{my_2 + ny_1}{m + n} \)
Given that \( x_1 = 7, y_1 = -6, x_2 = 3, y_2 = 4, m = 1, n = 2 \)
\( \therefore x = \frac{1(3) + 2(7)}{1 + 2} = \frac{3 + 14}{3} = \frac{17}{3} \)
\( y = \frac{1(4) + 2(-6)}{1 + 2} = \frac{4 - 12}{3} = \frac{-8}{3} \)
As x-coordinate is positive and y-coordinate is negative:
\( \therefore (x, y) = \left( \frac{17}{3}, \frac{-8}{3} \right) \) lies in the IV quadrant

Question. The distance between the points (a cos \( \theta \) + b sin \( \theta \), 0) and (0, a sin \( \theta \) – b cos \( \theta \)), is
(a) \( a^2 + b^2 \)
(b) \( a^2 - b^2 \)
(c) \( \sqrt{a^2 + b^2} \)
(d) \( \sqrt{a^2 - b^2} \)
Answer: (c) \( \sqrt{a^2 + b^2} \)
Explanation: Distance between the given points
= \( \sqrt{(0 - (a \cos \theta + b \sin \theta))^2 + ((a \sin \theta - b \cos \theta) - 0)^2} \)
= \( \sqrt{(a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \cos \theta \sin \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta)} \)
= \( \sqrt{a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta)} \)
= \( \sqrt{a^2 + b^2} \)

Question. The point which lies on the perpendicular bisector of the line segment joining point A (–2, –5) and B (2, 5) is:
(a) (0, 0)
(b) (0, –1)
(c) (–1, 0)
(d) (1, 0)
Answer: (a) (0, 0)

Question. The fourth vertex D of a parallelogram ABCD whose three vertices are A (–2, 3), B (6, 7) and C (8, 3) is:
(a) (0, 1)
(b) (0, –1)
(c) (–1, 0)
(d) (1, 0)
Answer: (b) (0, –1)
Explanation: It is given that ABCD is a parallelogram with vertices A (–2, 3), B (6, 7) and C (8, 3). Let fourth vertex be D (x, y). We know that diagonals AC and BD will bisect each other.
Midpoint of diagonal AC \( (x_1, y_1) \)
= \( \left( \frac{-2 + 8}{2}, \frac{3 + 3}{2} \right) = \left( \frac{6}{2}, \frac{6}{2} \right) = (3, 3) \)
Midpoint of diagonal BD \( (x_2, y_2) = \left( \frac{x + 6}{2}, \frac{y + 7}{2} \right) \)
But the two midpoints are the same. So,
\( \frac{x + 6}{2} = 3 \) and \( \frac{y + 7}{2} = 3 \)

\( \implies x + 6 = 6 \) and \( y + 7 = 6 \)

\( \implies x = 0 \) and \( y = -1 \)
Hence, the fourth vertex D (x, y) = (0, –1).

Question. If the point P(k, 0) divides the line segment joining the points A(2, –2) and B(–7, 4) in the ratio 1 : 2, then the value of k is
(a) 1
(b) 2
(c) –2
(d) –1
Answer: (d) –1
Explanation: Let point \( P(k, 0) \) divide \( A(2, -2) \) and \( B(-7, 4) \) in ratio \( 1 : 2 \).
Here, \( P(k, 0) = \left( \frac{1(-7) + 2(2)}{1 + 2}, \frac{1(4) + 2(-2)}{1 + 2} \right) \)
[By Section Formula]
\( P(k, 0) = \left( \frac{-7 + 4}{3}, \frac{4 - 4}{3} \right) \)
\( P(k, 0) = (-1, 0) \)

\( \implies k = -1 \)

Question. The distance of the point P(–3, –4) from the x-axis (in units) is:
(a) 3
(b) –3
(c) 4
(d) 5
Answer: (c) 4

Question. If the point P(2, 1) lies on the line segment joining points A(4, 2) and B(8, 4), then:
(a) \( AP = \frac{1}{3} AB \)
(b) \( AP = PB \)
(c) \( PB = \frac{1}{3} AB \)
(d) \( AP = \frac{1}{2} AB \)
Answer: (d) \( AP = \frac{1}{2} AB \)
Explanation: It is given that P(2, 1) lies on the line segment joining the points A(4, 2) and B(8, 4).
We know that distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \),
\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Distance between A(4, 2) and P(2, 1)
\( AP = \sqrt{(2 - 4)^2 + (1 - 2)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \)
Distance between A (4, 2) and B (8, 4)
\( AB = \sqrt{(8 - 4)^2 + (4 - 2)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \)
\( \therefore AB = 2\sqrt{5} = 2(AP) \)
\( \implies AP = \frac{AB}{2} \)

Question. If \( A \left(\frac{m}{3}, 5\right) \) is the mid-point of the line segment joining the points Q(–6, 7) and R(–2, 3), then the value of m is:
(a) –12
(b) –4
(c) 12
(d) –6
Answer: (a) –12
Explanation: Mid point is \( A \left(\frac{m}{3}, 5\right) \)

\( \implies \frac{m}{3} = \frac{-6 - 2}{2} \)

\( \implies \frac{m}{3} = -4 \)

\( \implies m = -12 \)

Question. The perimeter of a triangle ABC with vertices A (0, 4), B (0, 0) and C (3, 0) is:
(a) 5 units
(b) 11 units
(c) 12 units
(d) \( (7 + \sqrt{5}) \) units
Answer: (c) 12 units
Explanation: Perimeter of triangle = AB + BC + CA
\( AB = \sqrt{(0 - 0)^2 + (4 - 0)^2} = 4 \)
\( BC = \sqrt{(3 - 0)^2 + (0 - 0)^2} = 3 \)
\( CA = \sqrt{(3 - 0)^2 + (0 - 4)^2} = \sqrt{9 + 16} = 5 \)
Perimeter = 4 + 3 + 5 = 12 units

Question. If \( P \left(\frac{a}{3}, 4\right) \) is the midpoint of the line segment joining the points Q(–6, 5) and R(–2, 3), then the value of a is:
(a) –4
(b) –12
(c) 12
(d) –6
Answer: (b) –12
Explanation: Midpoint of QR = \( P \left( \frac{-6 - 2}{2}, \frac{5 + 3}{2} \right) = P(-4, 4) \)
Comparing with \( P \left(\frac{a}{3}, 4\right) \):
\( \frac{a}{3} = -4 \)
\( \implies a = -12 \)

Question. The perpendicular bisector of the line segment joining the points A(1, 5) and B(4, 6) cuts the y-axis at:
(a) (0, 13)
(b) (0, –13)
(c) (0, 12)
(d) (13, 0)
Answer: (a) (0, 13)

Question. A circle drawn with origin as the centre passes through \( \left(\frac{13}{2}, 0\right) \). The point which does not lie in the interior of the circle is:
(a) \( \left(\frac{-3}{4}, 1\right) \)
(b) \( \left(2, \frac{7}{3}\right) \)
(c) \( \left(5, \frac{-1}{2}\right) \)
(d) \( \left(-6, \frac{5}{2}\right) \)
Answer: (d) \( \left(-6, \frac{5}{2}\right) \)
Explanation: Radius of circle = Distance between (0, 0) and \( \left(\frac{13}{2}, 0\right) = 6.5 \)
Check distance of (d) from (0, 0):
\( \sqrt{(-6 - 0)^2 + \left(\frac{5}{2} - 0\right)^2} = \sqrt{36 + 6.25} = \sqrt{42.25} = 6.5 \)
Since the distance is equal to the radius, the point lies on the circle and not in the interior.

Fill in the Blanks

Question. AOBC is a rectangle whose three vertices are A(0, –3), O(0, 0) and B(4, 0). The length of its diagonal is .............................. .
Answer: 5

Question. The centroid of the triangle whose vertices are (4, – 8), (– 9, 7) and (8, 13) is .............................. .
Answer: (1, 4)
Explanation: Centroid \( = \left( \frac{4 - 9 + 8}{3}, \frac{-8 + 7 + 13}{3} \right) = \left( \frac{3}{3}, \frac{12}{3} \right) = (1, 4) \).

Question. The ratio in which x-axis divides the line segment joining the point (2, 3) and (4, – 8) is .............................. .
Answer: 3 : 8
Explanation: Let ratio be k : 1. Point on x-axis is (x, 0).
\( 0 = \frac{k(-8) + 1(3)}{k + 1} \)
\( \implies -8k + 3 = 0 \)
\( \implies k = \frac{3}{8} \). The ratio is 3:8.

Question. The mid-point of the line segment AB is (4, 0). If the co-ordinates of point A is (3,–2), then co-ordinates of point B is .............................. .
Answer: B (5, 2)
Explanation: Let B be (x, y).
\( \frac{3 + x}{2} = 4 \)
\( \implies x = 5 \)
\( \frac{-2 + y}{2} = 0 \)
\( \implies y = 2 \)
Hence, B is (5, 2).

Question. Distance of a point (–24, 7) from the origin (in units) is .............................. .
Answer: 25 units
Explanation: \( d = \sqrt{(-24 - 0)^2 + (7 - 0)^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \).

Question. If P(–1, 1) is the mid-point of the line segment joining the points A(–3, b) and B(1, b + 4) then b = ...............................
Answer: –1
Explanation: \( 1 = \frac{b + (b + 4)}{2} \)
\( \implies 2 = 2b + 4 \)
\( \implies 2b = -2 \)
\( \implies b = -1 \).

Write True or False

Question. \( \Delta ABC \) with vertices A(–2, 0), B(2, 0) and C(0, 2) is similar to \( \Delta DEF \) with vertices D(–4, 0), E(4, 0) and F(0, 4).
Answer: True.
Explanation: Sides of \( \Delta ABC \) are \( AB = 4, BC = 2\sqrt{2}, CA = 2\sqrt{2} \). Sides of \( \Delta DEF \) are \( DE = 8, EF = 4\sqrt{2}, FD = 4\sqrt{2} \). Since sides are proportional \( \left( \frac{1}{2} \text{ ratio} \right) \), triangles are similar.

Question. Point P (–4, 2) lies on the line segment joining the points A (–4, 6) and B (–4, –6).
Answer: True.
Explanation: All three points have the same x-coordinate \( x = -4 \). The y-coordinate 2 lies between –6 and 6.

Question. Points A(4, 3), B(6, 4), C(5, –6) and D(–3, 5) are the vertices of a parallelogram.
Answer: False.
Explanation: Opposite sides are not equal. \( AB = \sqrt{5} \) but \( CD = \sqrt{121 + 5} = \sqrt{126} \). (Wait, OCR correction: \( CD = \sqrt{(5 - (-3))^2 + (-6 - 5)^2} = \sqrt{8^2 + (-11)^2} = \sqrt{64 + 121} \)). Opposite sides are not equal, so it's not a parallelogram.

Question. Point P(5, –3) is one of the two points of trisection of the line segment joining points A(7, –2) and B(1, –5).
Answer: True.
Explanation: Using section formula for 1:2 ratio: \( x = \frac{1(1) + 2(7)}{3} = \frac{15}{3} = 5 \) and \( y = \frac{1(-5) + 2(-2)}{3} = \frac{-9}{3} = -3 \). This matches P(5, –3).

Question. Point P(–2, 4) lies on a circle of radius 6 and centre C(3, 5).
Answer: False.
Explanation: \( CP = \sqrt{(3 - (-2))^2 + (5 - 4)^2} = \sqrt{5^2 + 1^2} = \sqrt{26} \). Since \( \sqrt{26} \neq 6 \), it doesn't lie on the circle.

Question. The points A (–1, –2), B (4, 3), C (2, 5) and D (–3, 0) in that order from a rectangle.
Answer: True.
Explanation: Opposite sides are equal \( AB = CD = \sqrt{50} \) and \( BC = AD = \sqrt{8} \). Diagonals are also equal \( AC = BD = \sqrt{58} \).