MCQ on Coordinate Geometry for Class 10 with Answers – Maths Class 10 MCQ Online Test are covered in this Article. Coordinate Geometry Class 10 MCQ Test contains 25 questions. Answers to MCQ on Coordinate Geometry for Class 10 are available after clicking on the answer. MCQ Questions for Class 10 with Answers have been made for Class 10 students to help check the concept you have learnt from detailed classroom sessions and application of your knowledge.
| Board | CBSE |
| Textbook | Maths (NCERT) |
| Class | Class 10 |
| Chapter | Chapter 7 Coordinate Geometry |
| Category | MCQ Questions for Class 10 Maths with Answers |
MCQ on Coordinate Geometry for Class 10 with Answers
1. The coordinates of a point on the x-axis are of the form
(a) (x,0)
(b) (0, x)
(c) (0,0)
(d) None of these
Answer
Answer: (a) (x,0)
2.The distance between the points A (x1, y1) and B (x2, y2) is
(a)√(x2-x1)2 – (y2-y1)2
(b) √(x2-x1)2 + (y2-y1)2
(c) √(x2-x1)2 (y2-y1)2
(d) √(x2-x1)2
Answer
Answer: (b) √(x2-x1)2 + (y2-y1)2
Explanation: The distance between the points A (x1, y1) and B (y1, y2) = AB = √(x2-x1)2 +(y2-y1)2
3. Find the value of x such that the point (x, 2) is equidistant from the points (5, 3) and (6, 1).
(a)x=5.5
(b)x=5
(c)x=4
(d)x=4.5
Answer
Answer: (a)x=5.5
Explanation:
√(x-5)2 +(2-3)2 = √(x-6)2 +(2-1)2
(x-5)2 +(2-3)2 = (x-6)2 +(2-1)2
(x-5)2 +1 = (x-6)2 +1
x2 – 10x + 25 = x2 – 12x + 36
2x = 11
x = 5.5
4. The distance of a point P (4, 3) from the origin O (0, 0) is
(a)4
(b)3
(c)5
(d)6
Answer
Answer: (c)5
Explanation:
√(x2-x1)2 +(y2-y1)2 = √(0-4)2 +(0-3)2
= √(4)2 +(3)2
= √16+9
=√25
= 5
5. Find the values of y for which the distance between the points P (-10, –2) and Q (0, y) is 10 units
(a) 0
(b) -2
(c) 2
(d) 10
Answer
Answer: (b) -2
Explanation:
√(x2-x1)2 +(y2-y1)2 = PQ
√(0+10)2 +(y+2)2 = 10
100 + (y+2)2 = 100
(y+2)2=0
y = -2
6. Find a relation between x and y such that the point (x, y) is equidistant from the point (1, 5) and (– 2, 3).
(a) 2x + 4y = 13
(b) x + 2y = 14
(c) x + y=7
(d) x – 2y = 7
Answer
Answer: (a) 2x + 4y = 13
Explanation:
√(x-1)2 +(y-5)2 = √(x+2)2 +(y-3)2
(x-1)2 +(y-5)2 = (x+2)2 +(y-3)2
x2 – 2x +1+ y2 – 10y +25 = x2 +4x+4 +y2 -6y +9
2x + 4y = 13
7. Find the value of y such that the point (0, y) is equidistant from the points (-1, -3) and (1, -1).
(a) y = 7/4
(b) y = -2
(c) y = -4/7
(d) y = 4/7
Answer
Answer: (b) y = -2
Explanation:
√(0+1)2 +(y+3)2 = √(0-1)2 +(y+1)2
(1)2 +(y+3)2 = (1)2 +(y+1)2
Y2 +6y +9 = y2 +2y +1
4y = -8
y = -2
MCQ on Coordinate Geometry for Class 10 with Answers
8. Distance between P (a-c, b-c) and Q (a+c, b+c) is
(a) 2√2c
(b) 2√2b
(c) 2√2a
(d) √2c
Answer
Answer: (a) 2√2c
Explanation:
PQ = √(x2-x1)2 +(y2-y1)2
PQ = √(a+c-a+c)2 +(b+c-b+c)2
= √(2c)2+(2c)2
=√4c2 + 4c2
= 2√2c
9.The points A (1,2) B (5,6) and C (9,10) are collinear
(a)True
(b)False
Answer
Answer: (a)True
Explanation:
AB = √(x2-x1)2 +(y2-y1)2
= √(5-1)2 +(6-2)2
= √16+16 = 4√2
BC = √(9-5)2 +(10-6)2
= √16+16 = 4√2
AC = √(9-1)2 +(10-2)2
= √64+64 = 8√2
AB + BC = 4√2+4√2
= 8√2 = AC
So, from this we can say that A, B and C are collinear.
10. the given points A (-1,-2), B (1, 0), C (-1, 2), D (-3,0) are vertices of
(a)square
(b) rectangle
(c) parallelogram
(d) None of these
Answer
Answer: (a)square
Explanation:
AB = √(x2-x1)2 +(y2-y1)2
= √(-1-1)2 +(-2-0)2
= √(-2)2 +(-2)2 = 2√2
BC = √(1+1)2 +(0-2)2
= √(2)2 +(-2)2 = 2√2
CD = √(-1+3)2 +(2-0)2
= √(2)2 +(2)2 =2√2
AD = √-3+1)2 +(0+2)2
= √(-2)2 +(2)2 = 2√2
AC (diagonal) = √(-1+1)2 +(-2-2)2
= √(0)2 +(4)2 = 4
BD (diagonal) = √(1+3)2 +(0-0)2
= √(4)2 +(0)2 = √16+0 = 4
So, this forms a square.
11.The coordinates of the mid-point P of the join of the points A (x1, y1) and B (x2, y2) is
(a)((x1 – x2)/2 , (y1 – y2)/2)
(b)((x1 – x2)/2 , (y1 + y2)/2)
(c)((x1 + x2)/2 , (y1 – y2)/2)
(d)((x1 + x2)/2 , (y1 + y2)/2)
Answer
Answer: (d)((x1 + x2)/2 , (y1 + y2)/2)
Ques12. The coordinates of the point P (x, y) which divides the line segment joining the points A (x1, y1) and B (x2, y2) internally in the ratio m1: m2 are
(a) ((m2 x1+m1 x2)/(m1+m2 ),(m2 y1+m1 y2)/(m1+m2 ))
(b) ((m2 x1+m1 x2)/(m1-m2 ),(m2 y1+m1 y2)/(m1 – m2 ))
(c) ((m2 x1 – m1 x2)/(m1+m2 ),(m2 y1 – m1 y2)/(m1+m2 ))
(d) ((m2 x1 – m1 x2)/(m1+m2 ),(m2 y1+m1 y2)/(m1-m2 ))
Answer
Answer: (a) ((m2 x1+m1 x2)/(m1+m2 ),(m2 y1+m1 y2)/(m1+m2 ))
13. Find the coordinates of the point which divides the join of (–3, 10) and (2, 1) in the ratio 2:3.
(a) (1,6.5)
(b) (-1,6.4)
(c) (-1, -6.5)
(d) (1, -6.5)
Answer
Answer: (b) (-1,6.4)
Explanation:
Using section formula,
x = (m2 x1+m1 x2)/(m1+m2 )
= (3(-3)+2(2))/(2+3)
= (-9+4)/5
= (-5)/5 = -1
y = (m2 y1+m1 y2)/(m1+m2 )
= (3(10)+2(1))/(2+3)
= (30+2)/5
= 32/5 = 6.4
14. Find the ratio in which the line segment joining the points A (9, 1) and B (-5, 7) is divided by (2, 4)
(a) 1:2
(b) 2:1
(c) 2:3
(d) 1:1
Answer
Answer: (d) 1:1
Explanation:
Let (2, 4) divide AB internally in the ratio m1: m2.
Using the section formula
(2, 4) = ((m2 x1+m1 x2)/(m1+m2 ),(m2 y1+m1 y2)/(m1+m2 ))
= ((9m2-5m1)/(m1+m2 ),(1m2+7m1)/(m1+m2 ))
2 = (9m2-5m1)/(m1+m2 )
2(m1+m2) = 9m2-5m1
2m2+2m1 = 9m2-5m1
7m2-7m1 = 0
m2=m1
m1 :m2 = 1:1
15.The coordinates of mid-point P of the join of the points A (a+2, b-2) and B (-a-2, b-2) is
(a) (a, b-2)
(b) (0, -2)
(c) (0, b)
(d) (0, b-2)
Answer
Answer: (d) (0, b-2)
Explanation:
Mid- point of AB = ((x1+x2)/2,(y1+y2)/2)
= ((a+2-a-2)/2,(b-2+b-2)/2)
= (0/2,(2b-4)/2)
= (0, b-2)
16. The point which divides the line segment joining the points (-7, 5) and (-3, 2) in ratio 1: 2 internally lies in the
(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant
Answer
Answer: (b) II quadrant
Explanation:
Using section formula,
x = (m2 x1+m1 x2)/(m1+m2 )
= (2(-7)+1(-3))/(1+2)
= (-14-3)/3 = (-17)/3
y = (m2 y1+m1 y2)/(m1+m2 )
= (2(5)+1(2))/(1+2)
= (10+2)/3
= 12/3 = 4
(-x, y ) lies in II quadrant.
MCQ on Coordinate Geometry for Class 10 with Answers
17. If (3, -2), (x, 4), (-2, y) and (0, 5) are the vertices of a parallelogram taken in order, find x and y
(a) x=11, y =1
(b) x=1, y =11
(c) x=-1, y =1
(d) x=11, y =-1
Answer
Answer: (b) x=1, y =11
Explanation:
We know that diagonals of a parallelogram bisect each other.
Mid point of AC = Mid point of BD
((3-2)/2,(-2+y)/2) = ((x+0)/2,(4+5)/2)
(1/2,(-2+y)/2) = (x/2,9/2)
By comparing,
x/2 = 1/2
x = 1
And, (-2+y)/2 = 9/2
y = 11
18. Find the coordinates of the centre of a circle whose diameter is AB and points are A(9,-8) and B(-3, 20)
(a) (3,6)
(b) (-3,6)
(c) (3, -6)
(d) None of these
Answer
Answer: (a) (3,6)
Explanation:
We know that centre is the mid-point of the diameter of a circle. Let the centre be (x, y)
(x, y) =( (9-3)/2,(-8+20)/2)
= (6/2,12/2)
= (3,6)
19. Find the area of a square if its vertices are A (1, 7), B (4, 2), C (–1, –1) and D (– 4, 4) taken in order
(a) 34 square units
(b) 17 square units
(c) 47 square units
(d) 27 square units
Answer
Answer: (a) 34 square units
Explanation:
All sides of a square are equal. So, length of all sides will be equal.
AB = √(4-1)2 +(2-7)2
= √32 +(5)2 = √34
BC = √(4+1)2 +(2+1)2
= √52 +(3)2 = √34
CD =√(-4+1)2 +(4+1)2
= √32 +(5)2 = √34
AD =√(-4-1)2 +(4-7)2
= √52 +(3)2 = √34
Area of square ABCD = (side)2
= (√34)2 = 34 square units
20. If the distance between the points P (m, -2) and Q (0, 3) is 6, then the value of m is
(a) m= √11
(b) m= -√11
(c) m= ±√11
(d) None of these
Answer
Answer: (c) m= ±√11
Explanation:
√(x2-x1)2 +(y2-y1)2 = PQ
√(0-m)2 +(3+2)2 = 6
m2 + 25 =36
m2 = 11
m= ±√11
21. Find the coordinates of the points of trisection of the line segment joining the points
A (-1,3) and B (1, -4).
(a) P=(-1/3 , 2/3 ) and Q =(1/3 ,-5/3 )
(b) P=(1/3 , -2/3 ) and Q =(1/3 ,-2/3 )
(c) P=(-1/3 , 2/3 ) and Q =(-1/3 ,-2/3 )
(d) P=(-1/3 , -2/3 ) and Q =(-1/3 ,-2/3 )
Answer
Answer: (a) P=(-1/3 , 2/3 ) and Q =(1/3 ,-5/3 )
Explanation:
Let P and Q be the points of trisection of AB i.e., AP = PQ = QB
P divides AB internally in the ratio 1:2.
Using section formula,
Coordinates of P = ((m2 x1+m1 x2)/(m1+m2 ),(m2 y1+m1 y2)/(m1+m2 ))
= ((2(-1)+1(1))/(1+2),(2(3)+1(-4))/(1+2))
= ((-2+1)/3,(6-4)/3)
= (-1/3,2/3)
Q also divides AB internally in the ratio 2:1.
So, the coordinates of Q = ((m2 x1+m1 x2)/(m1+m2 ),(m2 y1+m1 y2)/(m1+m2 ))
= ((1(-1)+2(1))/(2+1),(1(3)+2(-4))/(2+1))
= ((-1+2)/3,(3-8)/3)
= (1/3,(-5)/3 )
22. Find the ratio in which the x-axis divides the line segment joining the points P (-2, -5) and Q (1,3)
(a) 1:1
(b) 5:3
(c) 3:1
(d) 5:1
Answer
Answer: (b) 5:3
Explanation:
Let the ratio be k: 1.
Using section formula,
The coordinates of the point which divides PQ in the ratio k:1= ((m2 x1+m1 x2)/(m1+m2 ),(m2 y1+m1 y2)/(m1+m2 ))
= ((1(-2)+k(1))/(k+1),(1(-5)+3k)/(k+1))
=(k-2)/(k+1),(3k-5)/(k+1))
This point lies on the x-axis, and we know that on the x-axis the ordinate is 0.
(3k-5)/(k+1)=0
3k-5 = 0
k = 5/3
The ratio is 5/3 ∶1 or 5:3
23. Find the point on the y-axis which is equidistant from (√5, √2 ) and (√3 , 9)
(a) (0, 77/(9-√2)2)
(b) (77/(9-√2)2, 0)
(c) (0, -77/(9-√2)2)
(d) None of these
Answer
Answer: (a) (0, 77/(9-√2)2)
Explanation:
This point lies on the y-axis, and we know that on the y-axis the abscissa is 0.
√(0-√5)2 +(y-√2)2 = √(0-√3)2 +(y-9)2
√(√5)2 +(y-√2)2 = √(√3)2 +(y-9)2
5 + y2 -2√2y +2 = 3 + y2 -18y +81
18y -2√2y = 84 – 7
2y (9-√2) = 77
y = 77/((9-√2)2)
24. Distance between the points A (x,0) and B (y,2) is
(a)√( y-x)2 + 4
(b)√( y-x)2 – 4
(c)√( y+x)2 + 4
(d)√( y+x)2 – 4
Answer
Answer: (a)√( y-x)2 + 4
Explanation:
AB = √(y-x)2 +(2-0)2
= √(y-x)2 +4
25. Find the coordinates of the point which divides the line segment joining the points (-1, 2) and (0, -5) in the ratio 4: 2 internally.
(a) (1/3, 8/3 )
(b) (1/3, -8/3 )
(c) (-1/3, 8/3 )
(d) (-1/3, -8/3 )
Answer
Answer: (d) (-1/3, -8/3 )
Explanation:
Coordinates of the point which divides the line segment joining the points (-1, 2) and (0, -5) = ((2(-1)+4(0))/(4+2),(2(2)+4(-5))/(4+2))
= ((-2+0)/6,(4-20)/6)
= ((-2)/6,(-16)/6)
= ((-1)/3,(-8)/3)
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Frequently Asked Questions on Coordinate Geometry for Class 10 with Answers
1. Are these MCQ on Coordinate Geometry for Class 10 are based on 2021-22 CBSE Syllabus?
Yes . There are 25 MCQ’s on this Chapter in this blog.
2. Are you giving all the chapters of Maths Class 10 MCQs with Answers which are given in CBSE syllabus for 2021-22 ?
Yes, we are providing all the chapters of Maths Class 10 MCQs with Answers.