# Class 10 Maths Chapter 7 Coordinate Geometry MCQs

Class 10 Maths MCQs for Chapter 7 (coordinate geometry) are provided here, online. Students can practice these multiple-choice questions, which are prepared as per the CBSE syllabus and NCERT curriculum. It will help students to score good marks in the board exam. All the objective questions are presented here with their answers and detailed explanations.

## Class 10 Maths MCQs for Coordinate Geometry

Get the MCQs for coordinate geometry to understand the concept well. Students are suggested to practice these questions by themselves and then verify the answers. Get important questions for class 10 Maths here as well.

MCQs for Coordinate Geometry

1.The points (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0) forms a quadrilateral of type:

(a)Square

(b)Rectangle

(c)Parallelogram

(d)Rhombus

Explanation: Let A(- 1, – 2), B(1, 0), C(- 1, 2) and D(- 3, 0) are the four vertices of quadrilateral.

By distance formula, we know:

$AB= \sqrt{(1+1)^{2} + (0+2)^{2}}= 2\sqrt{2}$

$BC= \sqrt{(-1-1)^{2} + (2-0)^{2}}= 2\sqrt{2}$

$CD= \sqrt{(-3+1)^{2} + (0-2)^{2}}= 2\sqrt{2}$

$DA= \sqrt{(-3+1)^{2} + (2)^{2}}= 2\sqrt{2}$

$AC= \sqrt{(-1+1)^{2} + (2+2)^{2}}= 4$

$BD= \sqrt{(1+3)^{2} + (0-0)^{2}}= 4$

Hence, the length of the sides = 2√2

Diagonals = 4

Hence, the given points form a square.

2.If the distance between the points A(2, -2) and B(-1, x) is equal to 5, then the value of x is:

(a)2

(b)-2

(c)1

(d)-1

Explanation: By distance formula, we know:

\begin{aligned} &\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}=5\\ &\sqrt{(-1-2)^{2}+(x+2)^{2}}=5\\ &\sqrt{9+(x+2)^{2}}=5 \end{aligned}

9+(x+2)2=25

(2+x)2=16

Take square root on both the sides,

2+x=4

x=2

3.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)

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.

4.The distance of point A(2, 4) from x-axis is

(a)2

(b)4

(c)-2

(d)-4

Explanation: Distance of a point from x-axis is equal to the ordinate of the point.

5.The distance between the points P(0, 2) and Q(6, 0) is

(a)4√10

(b)2√10

(c)√10

(d)20

Explanation: By distance formula we know:

$PQ=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

PQ = √[(6-0)2+(0-2)2]

PQ = √(62+22)

PQ=√(36+4)

PQ=√40=2√10

6.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

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.

7.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)

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)

8.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

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.

9.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)

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).

10.The area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order, is:

(a)12 sq.unit

(b)24 sq.unit

(c)30 sq.unit

(d)32 sq.unit

Explanation: To find the area of the rhombus, we need to find the length of its diagonals and use the below formula:

Area = ½ (Diagonal1)(Diagonal2)

Area = (1/2 ) (AC)(BD)

Diagonal1=√[(3-(-1))2+(0-4)2]= 4√2

Diagonal2 = √[(4-(-2))2+(5-(-1))2]=6√2

Area = ½ x 4√2 x 6√2 = 24 sq.unit.

#### 1 Comment

1. Satvik sharma

Good , their should more MCQ. These are so less