MCQ Questions for Class 10 Maths Coordinate Geometry with Answers
Class 10 Maths MCQs Chapter 7 Coordinate Geometry
1. The distance of the point P(2, 3) from the x-axis is
(a) 2
(b) 3
(c) 1
(d) 5
Answer/ Explanation
Answer: b
Explaination: Reason: The distance from x-axis is equal to its ordinate i.e., 3
2. The distance between the point P(1, 4) and Q(4, 0) is
(a) 4
(b) 5
(c) 6
(d) 3√3
Answer/ Explanation
Answer: b
Explaination: Reason: The required distance = \(\sqrt{(4-1)^{2}+(0-4)^{2}}=\sqrt{9+16}=\sqrt{25}=5\)
3. The points (-5, 1), (1, p) and (4, -2) are collinear if
the value of p is
(a) 3
(b) 2
(c) 1
(d) -1
Answer/ Explanation
Answer: d
Explaination: Reason: The points are collinear if area of Δ = 0
= \(\frac{1}{2}\)[-5(p + 2) +l(-2 -1) + 4(1 – p)] – 0
⇒ -5 p -10-3 + 4-4p = 0
⇒ -9p = +9
∴ p = -1
4. The area of the triangle ABC with the vertices A(-5, 7), B(-4, -5) and C(4, 5) is
(a) 63
(b) 35
(c) 53
(d) 36
Answer/ Explanation
Answer: c
Explaination: Reason: Area of ΔABC = \(\frac{1}{2}\) [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = \(\frac{1}{2}\)[-5(-5 – 5) -4(5 – 7) + 4(7 – (-5))] = \(\frac{1}{2}\)[-5(-10) -4(-2) + 4(12)] = \(\frac{1}{2}\)[50 + 8 + 48] = \(\frac{1}{2}\) × 106 = 53 sq. units
5. The distance of the point (α, β) from the origin is
(a) α + β
(b) α² + β²
(c) |α| + |β|
(d) \(\sqrt{\alpha^{2}+\beta^{2}}\)
Answer/ Explanation
Answer: d
Explaination: Reason: Distance of (α, β) from origin (0, 0) = \(\sqrt{(\alpha-0)^{2}+(\beta-0)^{2}}=\sqrt{\alpha^{2}+\beta^{2}}\)
6. The area of the triangle whose vertices are A(1, 2), B(-2, 3) and C(-3, -4) is
(a) 11
(b) 22
(c) 33
(d) 21
Answer/ Explanation
Answer: a
Explaination: Reason: Required area= \(\frac{1}{2}\)[1(3 + 4) -2(-4 – 2) -3(2 – 3)] = \(\frac{1}{2}\)[7 + 12 + 3] = \(\frac{1}{2}\) × 22 = 11
7. The line segment joining the points (3, -1) and (-6, 5) is trisected. The coordinates of point of trisection are
(a) (3, 3)
(b) (- 3, 3)
(c) (3, – 3)
(d) (-3,-3)
Answer/ Explanation
Answer: b
Explaination: Reason: Since the line segment AB is trisected
8. The line 3x + y – 9 = 0 divides the line joining the points (1, 3) and (2, 7) internally in the ratio
(a) 3 : 4
(b) 3 : 2
(c) 2 : 3
(d) 4 : 3
Answer/ Explanation
Answer: a
Explaination: Reason: Let the line 3x + y – 9 = 0 divide the line segment joining A(l, 3) ad B(2, 7) in the ratio K : 1 at point C.
9. The distance between A (a + b, a – b) and B(a – b, -a – b) is
Answer/ Explanation
Answer: c
Explaination:
10. If (a/3, 4) is the mid-point of the segment joining the points P(-6, 5) and R(-2, 3), then the value of ‘a’ is
(a) 12
(b) -6
(c) -12
(d) -4
Answer/ Explanation
Answer: c
Explaination:
11. If the distance between the points (x, -1) and (3, 2) is 5, then the value of x is
(a) -7 or -1
(b) -7 or 1
(c) 7 or 1
(d) 7 or -1
Answer/ Explanation
Answer: d
Explaination: Reason: We have \(\sqrt{(x-3)^{2}+(-1-2)^{2}}=5\)
⇒ (x – 3)² + 9 = 25
⇒ x² – 6x + 9 + 9 = 25
⇒ x² -6x – 7 = 0
⇒ (x – 7)(x + 1) = 0
⇒ x = 7 or x = -1
12. The points (1,1), (-2, 7) and (3, -3) are
(a) vertices of an equilateral triangle
(b) collinear
(c) vertices of an isosceles triangle
(d) none of these
Answer/ Explanation
Answer: b
Explaination: Reason: Let A(1, 1), B(-2, 7) and C(3, 3) are the given points, Then, we have
13. The coordinates of the centroid of a triangle whose vertices are (0, 6), (8,12) and (8, 0) is
(a) (4, 6)
(b) (16, 6)
(c) (8, 6)
(d) (16/3, 6)
Answer/ Explanation
Answer: d
Explaination: Reason: The co-ordinates of the centroid of the triangle is
14. Two vertices of a triangle are (3, – 5) and (- 7,4). If its centroid is (2, -1), then the third vertex is
(a) (10, 2)
(b) (-10,2)
(c) (10,-2)
(d) (-10,-2)
Answer/ Explanation
Answer: c
Explaination: Reason: Let the coordinates of the third vertex be (x, y)
15. The area of the triangle formed by the points A(-1.5, 3), B(6, -2) and C(-3, 4) is
(a) 0
(b) 1
(c) 2
(d) 3/2
Answer/ Explanation
Answer: a
Explaination: Reason: Area of ΔABC = \(\frac{1}{2}\) [-1.5(-2 – 4) + 6(4 – 3) + (-3) (3 + 2)] = \(\frac{1}{2}\) [9 + 6 – 15] = 0. It is a straight line.
16. If the points P(1, 2), B(0, 0) and C(a, b) are collinear, then
(a) 2a = b
(b) a = -b
(c) a = 2b
(d) a = b
Answer/ Explanation
Answer: a
Explaination: Reason: Area of ΔPBC = 0
⇒ \(\frac{1}{2}\)[1(0 – b) + 0(6 – 1) + a(2 – 0)] = 0
⇒ \(\frac{1}{2}\)[-6 + 2a] = 0
⇒ -b + 2a = 0
∴ 2a = b
17. A triangle with vertices (4, 0), (- 1, – 1) and (3, 5) is a/an
(a) equilateral triangle
(b) right-angled triangle
(c) isosceles right-angled triangle
(d) none of these
Answer/Explanation
Answer: a
Explaination:
18. The points (- 4, 0), (4, 0) and (0, 3) are the vertices of a/an [NCERT Exemplar Problems] (a) right triangle
(b) isosceles triangle
(c) equilateral triangle
(d) scalene triangle
Answer/Explanation
Answer: b
Explaination:
19. 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 [NCERT Exemplar Problems] 
Answer/Explanation
Answer: d
Explaination:
20. If the distance between the points(4, p) and (1, 0) is 5 units, then the value of p is [NCERT Exemplar Problems] (a) 4 only
(b) ± 4
(c) -4 only
(d) 0
Answer/Explanation
Answer: b
Explaination:
21. The points (2, 5), (4, – 1) and (6, – 7) are vertices of an/a
(a) isosceles triangle
(b) equilateral triangle
(c) right-angled triangle
(d) none of these
Answer/Explanation
Answer: d
Explaination:
22. If the segment joining the points (a, b) and (c, d) subtends a right angle at the origin, then
(a) ac – bd = 0
(b) ac + bd = 0
(c) ab + cd = 0
(d) ab – cd= 0
Answer/Explanation
Answer: b
Explaination:
Let A {a, b), B(c, d), 0(0, 0)
∴ ∠AOB = 90°
⇒ AB² = AO² + BO²
(c – a)² + (d- b)² = a² + b² + c² + d²
⇒ ac + bd = 0
23. AOBC is a rectangle whose three vertices are A(0, 3), 0(0, 0) and B(5, 0). The length of its diagonal is [NCERT Exemplar Problems] (a) 5
(b) 3
(c) √34
(d) 4
Answer/Explanation
Answer: c
Explaination:
24. The perimeter of a triangle with vertices (0,4), (0, 0) and (3, 0) is [NCERT Exemplar Problems] (a) 5
(b) 12
(c) 11
(d) 7 + √5
Answer/Explanation
Answer: b
Explaination:
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25. If the distance between the points (4, p) and (1, 0) is 5 units, then the value of p is [NCERT Exemplar Problems] (a) 4 only
(b) ±4
(c) -4 only
(d) 0
Answer/Explanation
Answer: b
Explaination:
26. If p\(\left(\frac{a}{3}, 4\right)\) is the mid-point of the line segment joining the points Q (-6, 5) and R (-2, 3), then the value of a is [NCERT Exemplar Problems] (a) -4
(b) -12
(c) 12
(d) -6
Answer/Explanation
Answer: b
Explaination:
27. If P(l, 2), Q(4, 6), R(5, 7) and S(a, b) are the vertices of a parallelogram PQRS, then
(a) a = 2,b = A
(b) a = 3,b = 4
(c) a = 2, b = 3
(d) a = 3, b = 5
Answer/Explanation
Answer: c
Explaination:
28. The point P which divides the line segment joining the points A(2, -5) and B(5, 2) in the ratio 2 : 3 lies in the quadrant
(a) I
(b) II
(c) III
(d) IV
Answer/Explanation
Answer: d
Explaination:|
29. The points (k + 1, 1), (2k + 1, 3) and (2k + 2, 2k) are collinear if
(a) k = -1, 2
(b) k=\(\frac{1}{2}\),2
(c) k = 2, 1
(d) k = –\(\frac{1}{2}\),2
Answer/Explanation
Answer: d
Explaination:
∵ Points are collinear.
∴ (k + 1) (3 – 2k) + (2k + 1) (2k- 1) + (2k + 2) (1 – 3) = 0
⇒ 3k + 3 – 2k² – 2k + 4k² – 1 -4k – 4 = 0
⇒ 2k² – 3k – 2 = 0
⇒ 2k² – 4k + k – 2 = 0
⇒ 2k(k – 2) + 1(k – 2) = 0
⇒ (2k + 1) (k – 2) = 0
k = –\(\frac{1}{2}\), k = 2
30. The area of the triangle with vertices at the points (a, b + c), (b, c + a) and (c, a + b) is
(a) (a + b + c) sq. units
(b) (a + b – c) sq. units
(c) (a – b + cj sq. units
(d) 0
Answer/Explanation
Answer: d
Explaination:
Using formula for area of triangle, we get
Area = zero.
Area of triangle
= \(\frac{1}{2}\)| a(c + a- a-b) + b(a+ b- b-c) + c(b + c- c-a) |
= \(\frac{1}{2}\) |ac – ab + ba – bc + cb – ca| = 0
31. The area (in square units) of the triangle formed by the points A(a, 0), 0(0, 0) and B(0, b) is
Answer/Explanation
Answer: b
Explaination:
A (a, 0), O(0, 0) and B(0, b)
ar(∆AOB) = \(\frac{1}{2}\)|a(0 – b) + 0(b – 0) + 0(0 -0)|
= \(\frac{1}{2}\)|-ab + 0 + 0| = \(\frac{ab}{2}\)
32. The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure is [NCERT Exemplar Problems] 
Answer/Explanation
Answer: a
Explaination:
∵ AOB is a right triangle.
∴ Mid-point of AB is equidistant from A, O and B.
33. The vertices of a triangle are (0, 0), (3, 0) and (0, 4). The centroid of the triangle is
Answer/Explanation
Answer: b
Explaination:
34. If A = (a , 2a) and B = \(\left(\frac{1}{a^{2}},-\frac{2}{a}\right)\) S = (1,0), then \(\frac{1}{\mathrm{SA}}+\frac{1}{\mathrm{SB}}=\) = _______ .
Answer/Explanation
Answer:
Explaination:
35. Find the distance of a point P(x, y) from the origin. [CBSE 2018] Answer:Answer/Explanation
Explaination:
36. Find the distance between the points \(\left(-\frac{8}{5}, 2\right) \text { and }\left(\frac{2}{5}, 2\right)\)
Answer/Explanation
Answer:
Explaination:
37. If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k? [Delhi 2017] Answer:Answer/Explanation
Explaination:
38. Show that (1, -1) is the centre of the circle circumscribing the triangle whose angular points are (4, 3), (- 2, 3) and (6, – 1).
Answer/Explanation
Answer:
Explaination:
39. A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6, 8) lies outside the circle. State whether true or false. Justify your answer. [NCERT Exemplar Problems] Answer:Answer/Explanation
Explaination:
True, because distance between centre (origin) and Q(6,8) is greater than its radius, i.e. 5.
40. Find the coordinates of a point A, where AB is diameter of a circle whose centre is (2, -3) and B is the point (1, 4). Answer:
[Delhi 2019] Answer/Explanation
Explaination:
AB is diameter of the circle.
Let C be centre of circle, coordinates of C are (2, -3). So, C is mid-point of AB (diameter).
Let coordinates of A are (x, y).
41. A straight line is drawn joining the points (3, 4) and (5, 6). If the line is extended, the ordinate of the point on the line, whose abscissa is -1 is ________ .
Answer/Explanation
Answer:
Explaination:
42. In figure, P(5, -3) and Q(3,y) are the points of trisection of the line segment joining A(7, -2) and B (1, -5). Then y equals ________ .
Answer/Explanation
Answer:
Explaination:
43. A(5, 1); B(l, 5) and C(-3, – 1) are the vertices of ∆ABC. Find the length of median AD. [CBSE 2018 (C)] Answer:Answer/Explanation
Explaination:
AD is median of ΔABC
∴ D is mid-point of BC
Coordinates of
44. If the mid-point of the line segment joining the points P(6, b – 2) and Q(- 2, 4) is (2, – 3), find the value of b.
Answer/Explanation
Answer:
Explaination:
45. If A(1, 2), B(4, 3) and C(6, 6) are the three vertices of a parallelogram ABCD, find the coordinates of the fourth vertex D.
Answer/Explanation
Answer:
Explaination:
46. Find the coordinates of the centroid of a triangle whose vertices are (0, 6), (8, 12) and (8, 0).
Answer/Explanation
Answer:
Explaination:
47. Two vertices of a triangle are (3, -5) and (-7, 4). If its centroid is (2, – 1), find the third vertex.
Answer/Explanation
Answer:
Explaination:
48. The coordinates of one end point of a diameter of a circle are (4, -1) and the coordinates of the centre are (1, -3). Find the coordinates of the other end of the diameter.
Answer/Explanation
Answer:
Explaination:
Given that coordinates of one end point of the diameter is (4, -1) and centre of the circle is (1, – 3).
Let coordinates of the other end of the diameter be (x, y).
We know that the centre of the circle (1, -3) is the mid-point of diameter.
⇒ \(\frac{4+x}{2}\) = 1 and \(\frac{(-1+y)}{2}\) ,
⇒ 4 + x = 2 and -1 + y = -6
⇒ x = – 2 and y = -6 + 1 = -5
Thus, coordinates of the other end of the diameter are (-2, -5).
49. Point P divides the line segment joining the points A(2, -5) and B(5, 2) in the ratio 2:3. Name the quadrant in which P lies. [Delhi 2011] Answer:Answer/Explanation
Explaination:
50. In figure, P(5, -3) and Q(3, vj are the points of trisection of the line segment joining A(7, -2) and B(1, -5). Find y [Foreign 2013] Answer:Answer/Explanation
Explaination:
51. Area of the triangle formed by (1, – 4), (3, – 2) and (- 3, 16) is _______ .
Answer/Explanation
Answer:
Explaination:
Area of the ∆ = \(\frac{1}{2}\)|1(-2 – 16) + 3(16 + 4) + (-3) (-4 + 2)|
= \(\frac{1}{2}\)| -18 + 60 + 6 | = 24 sq. units
52. If the points (- 2, – 5), (2, – 2) and (8, p) are collinear, then the value of p is _______ .
Answer/Explanation
Answer:
Explaination:
Points are collinear
∴ x1y2 + x2y3 + x3y1 – x2y1– x3y2 – x1y3 = 0
⇒ (-2) × (-2) + (2) × (p) + 8 × (-5) – (2) × (-5) – (8) × (-2) – (-2) ×(p) = 0
⇒ 4 + 2p – 40 + 10 + 16 + 2p = 0
⇒ 4p – 40 + 30 = 0
⇒ 4p = 10 10 5
⇒ p = \(\frac{10}{4}\) = \(\frac{5}{2}\)/
53. Show that the points P(\(\frac{-3}{2}\), 3), Q(6, -2) and R(-3, 4) are collinear.
Answer/Explanation
Answer:
Explaination:
If [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0 then points are collinear.
∴ \(\frac{-3}{2}\)(-2 – 4) + 6(4 – 3) + [(-3){3-(-2)}] = \(\frac{-3}{2}\) × (-6) + 6 × 1 – 3 × 5
= 9 + 6 – 15 = 0.
∴ P, Q and R are collinear.
54. Find the area (in square units) of the triangle formed by the points A(a, 0), 0(0, 0) and B(0, b). [Foreign 2011] Answer:Answer/Explanation
Explaination:
A (a, 0), O(0, 0) and B(0, b)
ar(∆AOB) = \(\frac{1}{2}\)|a(0 – b) + 0{b – 0) + 0(0 -0)|
= \(\frac{1}{2}\)| -ab + 0 +0| = \(\frac{ab}{2}\)
55. In figure, find the area of triangle ABC.
Answer/Explanation
Answer:
Explaination:
56. If area of the triangle given below is 20 square units, what are the coordinates of point C
Answer/Explanation
Answer:
Explaination:
Drow AD ⊥ OC
∴ AD = b
and coordinates of C are (h, 0)
∴ OC = h
Area of ∆OAC = 20 sq. units
⇒ \(\frac{1}{2}\) × OC × AD = 20
h × b = 40
h = \(\frac{40}{b}\)
∴ Coordinates of C are (\(\frac{40}{b}\), 0)
