# 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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