## Sunday, July 25, 2021

### Mathematics test class 10th │ Maths test │ Coordinate Geometry test

Mathematics test class 10th │ Maths test │ Coordinate Geometry test │ Class 10th Maths test

# 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

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

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

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

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) $$\sqrt{\alpha^{2}+\beta^{2}}$$

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

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)

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

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

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

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

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

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)

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)

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

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

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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