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 = \(\frac{1}{2}\) [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

= \(\frac{1}{2}\)[-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

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