MCQ Questions for Class 10 Maths Coordinate Geometry with Answers

MCQ Questions for Class 10 Maths Coordinate Geometry with Answers

Free PDF Download of CBSE Class 10 Maths Chapter 7 Coordinate Geometry Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Coordinate Geometry MCQs with Answers to know their preparation level.

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

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

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

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

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