This chapter introduces students to coordinate geometry in mathematics. Mathematics is understood to be a troublesome subject and this text is meant to assist students to work out the way to study the vital topics within the chapter by going over the MCQ questions that are provided by Sarthaks. This article shows students the way to steel oneself against any upcoming examinations by helping them practice the sums that are relevant and have a better chance of coming into the exam.

Mathematics may be a complex subject to review on your own and to assist students can use Class 10 Maths MCQ questions of coordinate geometry with answers provided to ace their examinations. Class 10 boards are vital for students and to assist prepare and had best they will do the questions as a neighborhood of their revision.

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Explore various MCQ Questions for Class 10 with answers provided with detailed solutions by looking below:-

**1. The horizontal and vertical lines drawn to determine the position of a point in a Cartesian plane are called**

(a) Intersecting lines

(b) Transversals

(c) Perpendicular lines

(d) X-axis and Y-axis

**2. 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

**3. The distance of the point P (2, 3) from the x-axis is**

(a) 2

(b) 3

(c) 1

(d) 5

**4. The points (3, 2), (0, 5), (-3, 2) and (0, -1) are the vertices of a quadrilateral. Which quadrilateral is it?**

(a) Rectangle

(b) Square

(c) Parallelogram

(d) Rhombus

**5. The distance between the point P(1, 4) and Q(4, 0) is**

(a) 4

(b) 5

(c) 6

(d) 3√3

**6. 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

**7. 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

**8. 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

**9. 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

**10. 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

**11. The distance of a point from the y axis is called its ———**

(a) Ordinate

(b) Abscissa

(c) Origin

(d) None of these

**12. The distance of a point from the x axis is called its —————**

(a) Ordinate

(b) Abscissa

(c) Origin

(d) None of these

**13. The midpoints of a line segment joining two points A(2, 4) and B(-2, -4)**

(a) (-2,4)

(b) (2,-4)

(c) (0, 0)

(d) (-2,-4)

**14. If O(p/3, 4) is the midpoint of the line segment joining the points P(-6, 5) and Q(-2, 3). The value of p is:**

(a) 7/2

(b) -12

(c) 4

(d) -4

**15. The points which divides the line segment of points P(-1, 7) and (4, -3) in the ratio of 2:3 is:**

(a) (-1, 3)

(b) (-1, -3)

(c) (1, -3)

(d) (1, 3)

**16. The ratio in which the line segment joining the points P(-3, 10) and Q(6, – 8) is divided by O(-1, 6) is:**

(a) 1:3

(b) 3:4

(c) 2:7

(d) 2:5

**17. The coordinates of a point P, where PQ is the diameter of circle whose centre is (2, – 3) and Q is (1, 4) is:**

(a) (3, -10)

(b) (2, -10)

(c) (-3, 10)

(d) (-2, 10)

**18. A linear equation in two variables of the form ax + by + c = 0, when represented graphically gives a ——————**

(a) Parabola

(b) Circle

(c) Straight line

(d) None of these

**19. The graph of a quadratic equation is a ———**

(a) Parabola

(b) Circle

(c) Straight line

(d) None of these

**20. 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

**Answers & Explanations **

1. Answer: (d) X-axis and Y-axis

2. Answer: (b) collinear

3. Answer: (b) 3

**Explanation:** The distance from x-axis is equal to its ordinate i.e., 3

4. Answer: (b) Square

5. Answer: (b) 5

**Explanation:** The required distance = \(\sqrt{(4-1)^2+(0-4)^2}\)

=\(\sqrt{9+16}\)

=\(\sqrt{25}\)

=5

6. Answer: (d) -1

**Explanation:** 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

7. Answer: (a) 11

**Explanation:** 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

8. Answer: (d) 7 or -1

**Explanation:** We have \(\sqrt{(x−3)^2+(−1−2)^2} =5\)

⇒ (x – 3)\(^2\) + 9 = 25

⇒ x\(^2\) – 6x + 9 + 9 = 25

⇒ x\(^2\) -6x – 7 = 0

⇒ (x – 7)(x + 1) = 0

⇒ x = 7 or x = -1

9. Answer: (a) 0

**Explanation:** 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.

10. Answer: (a) 2a = b

**Explanation:** 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

11. Answer: (b) Abscissa

12. Answer: (a) Ordinate

13. Answer: (c) (0, 0)

**Explanation:** As per midpoint formula, we know;

x=[2+(-2)]/2 = 0/2 = 0

y=[4+(-4)]/2=0/2=0

Hence, (0,0) is the midpoint of of AB.

14. Answer: (b)-12

**Explanation:** Since, (p/3, 4) is the midpoint of line segment PQ, thus;

p/3 = (-6-2)/2

p/3 = -8/2

p/3 = -4

p= -12

Therefore, the value of p is -12.

15. Answer: (d)(1, 3)

**Explanation:** By section formula we know:

x=[(2.4)+(3.(-1))]/(2+3) = (8-3)/5 = 1

y=[(2.(-3))+(3.7)]/(2+3) = (-6+21)/5 = 3

Hence, the required point is (1,3)

16. Answer: (c) 2:7

**Explanation:** Let the ratio in which the line segment joining P( -3, 10) and Q(6, -8) is divided by point O( -1, 6) be k :1.

So, -1 = ( 6k-3)/(k+1)

–k – 1 = 6k -3

7k = 2

k = 2/7

Hence, the required ratio is 2:7.

17. Answer: (a)(3, -10)

**Explanation:** By midpoint formula, we know;

[(x+1)/2,(y+4)/2] = (2,-3) (Since, O is the midpoint of PQ)

(x+1)/2 = 2

x+1=4

x=3

(y+4)/2 = -3

y+4=-6

y=-10

So, the coordinates of point P is (3, -10).

18. Answer: (c) Straight line

19. Answer: (a) Parabola

20. Answer: (d) 0

**Explanation:** 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

**Important Links: – MCQ Questions of Class 10 Maths all Chapter with Answers**

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