CBSE Class 10 Mathematics Coordinate Geometry MCQs Set K

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

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 \( \frac{y + 7}{2} = 3 \)
\( \Rightarrow x + 6 = 6 \) and \( y + 7 = 6 \)
\( \Rightarrow x = 0 \) 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:
Ratio 1:2, A(2, -2), B(-7, 4), P(k, 0)
Here, \( P(k, 0) = \left( \frac{1(-7) + 2(2)}{1 + 2}, \frac{1(4) + 2(-2)}{1 + 2} \right) = \left( \frac{-7 + 4}{3}, \frac{4 - 4}{3} \right) \), i.e., \( (-1, 0) \)
[By Section Formula]
\( \Rightarrow k = -1 \)

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) \( AP = \frac{1}{2} AB \)

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: A (– 3, b) and B (1, b + 4)
Mid point is \( \frac{m}{3} = \frac{-6 - 2}{2} \)
\( \Rightarrow \frac{m}{3} = -4 \)
\( \Rightarrow m = -12 \)

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
\( = \sqrt{(0-0)^2 + (4-0)^2} + \sqrt{(3-0)^2 + (0-0)^2} + \sqrt{(3-0)^2 + (0-4)^2} \)
\( = \sqrt{16} + \sqrt{9} + \sqrt{25} \)
= 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: It is given that \( P \left( \frac{a}{3}, 4 \right) \) is the midpoint of the line segment joining the points Q (–6, 5) and R (–2, 3).
We know that midpoint (x, y) \( = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
Midpoint of QR \( = P \left( \frac{-6 - 2}{2}, \frac{5 + 3}{2} \right) = \left( \frac{-8}{2}, \frac{8}{2} \right) \)
= P(–4, 4)
But mid-point is given as \( P \left( \frac{a}{3}, 4 \right) \) comparing the two, we get
\( P \left( \frac{a}{3}, 4 \right) = (-4, 4) \)
\( \Rightarrow \frac{a}{3} = -4 \Rightarrow a = -12 \)
Hence, the required value of a is –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: It is given that centre of the circle is origin O (0, 0) and it passes through \( \left( \frac{13}{2}, 0 \right) \).
\( \Rightarrow \) Radius of circle = Distance between (0, 0) and \( \left( \frac{13}{2}, 0 \right) \).
\( = \sqrt{\left( \frac{13}{2} - 0 \right)^2 + (0 - 0)^2} = \frac{13}{2} = 6.5 \)
We know that the point which does not lie in the interior of circle will be at a distance greater than the radius from the centre
(A) Distance between (0, 0) and \( \left( -\frac{3}{4}, 1 \right) \)
\( = \sqrt{\left( -\frac{3}{4} - 0 \right)^2 + (1 - 0)^2} = \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = \frac{5}{4} = 1.25 \)
Clearly, 1.25 < 6.5
\( \Rightarrow \) Point \( \left( -\frac{3}{4}, 1 \right) \) lies in the interior to the circle.
(B) Distance between (0, 0) and \( \left( 2, \frac{7}{3} \right) \)
\( = \sqrt{(2 - 0)^2 + \left( \frac{7}{3} - 0 \right)^2} = \sqrt{4 + \frac{49}{9}} = \sqrt{\frac{36 + 49}{9}} = \frac{\sqrt{85}}{3} = \frac{9.22}{3} = 3.1 \)
clearly 3.1 < 6.5
\( \Rightarrow \) Point \( \left( 2, \frac{7}{3} \right) \) lies in the interior of the circle.
(C) Distance between (0, 0) and \( \left( 5, -\frac{1}{2} \right) \)
\( = \sqrt{(5 - 0)^2 + \left( -\frac{1}{2} - 0 \right)^2} = \sqrt{25 + \frac{1}{4}} = \sqrt{\frac{101}{4}} = \frac{10.04}{2} = 5.02 \)
Clearly, 5.02 < 6.5.
\( \Rightarrow \) Point \( \left( 5, -\frac{1}{2} \right) \) lies in the interior of the circle.
(D) Distance between (0, 0) and \( \left( -6, \frac{5}{2} \right) \)
\( = \sqrt{(-6 - 0)^2 + \left( \frac{5}{2} - 0 \right)^2} = \sqrt{36 + \frac{25}{4}} = \sqrt{\frac{144 + 25}{4}} = \sqrt{\frac{169}{4}} = \frac{13}{2} = 6.5 \)
= 6.5 = radius
So the point \( \left( -6, \frac{5}{2} \right) \) lies on the circle and not in the interior.

Fill in the Blanks

Fill in the blanks/tables with suitable information:

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 of triangle having vertices \( (x_1, y_1), (x_2, y_2) \) and \( (x_3, y_3) \) is given by \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)
Required 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 (x, 0) be the point on x-axis divides the segment joining (2, 3) and (4, – 8) in ratio k:1.
\( 0 = \frac{-8k + 3}{k + 1} \Rightarrow 8k = 3 \Rightarrow k = \frac{3}{8} \)

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 (3, 2)
Explanation: Let coordinates of B be (x, y). Here A (3, – 2), B (x, y) and mid point (4, 0)
\( (4, 0) = \left( \frac{3 + x}{2}, \frac{-2 + y}{2} \right) \)
On comparing, we get \( 4 = \frac{3 + x}{2} \) and \( 0 = \frac{-2 + y}{2} \)
\( \Rightarrow 3 + x = 6 \) and \( -2 + y = 0 \)
\( \Rightarrow x = 3 \) and \( y = 2 \)
\( \therefore \) Coordinates of B is (3, 2)

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

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: A (– 3, b) and B (1, b + 4). Mid-point of AB = (– 1, 1)
\( (-1, 1) = \left( \frac{-3 + 1}{2}, \frac{b + b + 4}{2} \right) \)
\( \Rightarrow 1 = \frac{2b + 4}{2} \Rightarrow 2b + 4 = 2 \Rightarrow 2b = -2 \Rightarrow b = -1 \)

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.

Question. Point P (–4, 2) lies on the line segment joining the points A (–4, 6) and B (–4, –6).
Answer: True.
[Refer to graph on page 139]

Question. Points A(4, 3), B(6, 4), C(5, –6) and D(–3, 5) are the vertices of a parallelogram.
Answer: False.

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.

Question. Point P(–2, 4) lies on a circle of radius 6 and centre C(3, 5).
Answer: False.

Question. The points A (–1, –2), B (4, 3), C (2, 5) and D (–3, 0) in that order from a rectangle.
Answer: True.