MCQ Questions | Class 10 Maths Chapter 7 | Coordinate Geometry with Answers

MCQ | Coordinate Geometry Class 10 | Sat A

Check the below NCERT MCQ Questions for Class 10 Maths Chapter 7 Coordinate Geometry with Answers Pdf free download. MCQ Questions for Class 10 Maths with Answers were prepared based on the latest exam pattern. We have Provided Coordinate Geometry Class 10 Maths MCQs Questions with Answers to help students understand the concept very well.

Objective Question | NCERT Maths Class 10

NCERT Solutions Class 10 Maths
Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	10th
Subject:	Maths
Chapter:	7
Chapters Name:	Coordinate Geometry
Medium:	English

Ncert Solutions for Class 10 Maths | Objective Type Questions

You can refer to NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Coordinate Geometry | Class 10 Maths | NCERT Solutions

Q1.	The distance between the point P(1, 4) and Q(4, 0) is
	A. 4
	B. 5
	C. 6
	D. 3√3

Ans: 5

Q2.	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

Ans: 11

Q3.	The coordinates of the centre of a circle passing through (1, 2), (3, – 4) and (5, – 6) is:
	A. (11, – 2)
	B. (-2, 11)
	C. (11, 2)
	D. (2, 11)

Ans: (11, 2)

Q4.	The distance of the point (– 3, 4) from the origin is
	A. 25 units
	B. 1 unit
	C. 7 units
	D. 5 units

Ans: 5 units

Q5.	The distance of the point P (2, 3) from the x-axis is
	A. 2
	B. 3
	C. 1
	D. 5

Ans: 3

Q6.	The distance between the points (a, a) and (−√3a,√3a) is
	A. 3√2a units
	B. 2√2a units
	C. 2√2 units
	D. 2 units

Ans: 2√2a units

Q7.	The mid-point of the line segment joining the points A (-2, 8) and B (-6, -4) is
	A. (-4, -6)
	B. (2, 6)
	C. (-4, 2)
	D. (4, 2)

Ans: (-4, 2)

Q8.	The area of the triangle formed by joining the mid-points of the sides of the triangle, whose vertices are (0, -1), (2, 1) and (0, 3) is
	A. 4
	B. 2
	C. 3
	D. 1

Ans: 1

Q9.	The ordinate of a point is twice its abscissa. If its distance from the point (4,3) is √10, then the coordinates of the point are
	A. (1,2) or (3,6)
	B. (1,2) or (3,5)
	C. (2,1) or (3,6)
	D. (2,1) or (6,3)

Ans: (1,2) or (3,6)

Q10.	If (a, 0) , (0, b) and (x, y) are collinear, then
	A. ay + bx = ab
	B. ax + by = 1
	C. ax – by = ab
	D. ay – bx = 1

Ans: ay + bx = ab

Q11.	The distance of the point P(2, 3) from the x-axis is
	A. 2
	B. 3
	C. 1
	D. 5

Ans: 3
Explaination: Reason: The distance from x-axis is equal to its ordinate i.e., 3

Q12.	The distance between the point P(1, 4) and Q(4, 0) is
	A. 4
	B. 5
	C. 6
	D. 3√3

Ans: 5
Explaination: Reason: The required distance = \(\sqrt{(4-1)^{2}+(0-4)^{2}}=\sqrt{9+16}=\sqrt{25}=5\)

Q13.	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

Ans: -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

Q14.	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

Ans: 53
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

Q15.	The distance of the point (α, β) from the origin is
	A. α + β
	B. α² + β²
	C. \|α\| + \|β\|
	D. \(\sqrt{\alpha^{2}+\beta^{2}}\)

Ans: \(\sqrt{\alpha^{2}+\beta^{2}}\)
Explaination: Reason: Distance of (α, β) from origin (0, 0) = \(\sqrt{(\alpha-0)^{2}+(\beta-0)^{2}}=\sqrt{\alpha^{2}+\beta^{2}}\)

Q16.	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

Ans: 3 : 4

Q17.	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

Ans: -12

Q18.	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

Ans: collinear

Q19.	If (3,0), (2,a), and (b,6) are the vertices of ABC whose centroid is (2,5), then the values of a and b are
	A. a = 3, b = -9
	B. a = 0, b = 2
	C. a = 1, b = 9
	D. a = 9, b = 1

Ans: a = 9, b = 1

Q20.	The mid point of the line segment joining A(2a,4) and B(-2,3b) is M (1,2a + 1). The values of a and b are
	A. 2,3
	B. 1,1
	C. -2,-2
	D. 2,2

Ans: 2,2

CBSE Solutions | Ncert Solutions Maths Class 10

NCERT Solutions | 10th Maths Guide

NCERT | Class 10 Maths Solution

Ncert Solutions for Class 10 Maths

MCQ Questions | Class 10 Maths Chapter 7

MCQ Questions | Class 10 Maths Chapter 7 | Coordinate Geometry with Answers

MCQ | Coordinate Geometry Class 10 | Sat A

Objective Question | NCERT Maths Class 10

Ncert Solutions for Class 10 Maths | Objective Type Questions

Coordinate Geometry | Class 10 Maths | NCERT Solutions

CBSE Solutions | Ncert Solutions Maths Class 10

NCERT Solutions | 10th Maths Guide

NCERT | Class 10 Maths Solution

Ncert Solutions for Class 10 Maths

Ncert Solutions for Class 10 Maths Chapter 1

Ncert Solutions for Class 10 Maths Chapter 2

Ncert Solutions for Class 10 Maths Chapter 3

Ncert Solutions for Class 10 Maths Chapter 4

Ncert Solutions for Class 10 Maths Chapter 5

Ncert Solutions for Class 10 Maths Chapter 6

Ncert Solutions for Class 10 Maths Chapter 7

Ncert Solutions for Class 10 Maths Chapter 8

Ncert Solutions for Class 10 Maths Chapter 9

Ncert Solutions for Class 10 Maths Chapter 10

Ncert Solutions for Class 10 Maths Chapter 11

Ncert Solutions for Class 10 Maths Chapter 12

Ncert Solutions for Class 10 Maths Chapter 13

Ncert Solutions for Class 10 Maths Chapter 14

Ncert Solutions for Class 10 Maths Chapter 15

NCERT SOLUTIONS

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