Class 10 Maths Chapter 7 Coordinate Geometry MCQs
Class 10 Maths MCQs for Chapter 7 (Coordinate Geometry) are provided here, online. Students can practise these multiple-choice questions, which are prepared as per the CBSE syllabus (2021 – 2022) and NCERT curriculum. All the MCQ questions are formulated according to the latest exam pattern. 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. Download the PDF to get more MCQs here.
Click here: Class 10 Maths MCQs All Chapters
Class 10 Maths MCQs for Coordinate Geometry
CBSE board has released the datasheet for Class 10 students. It is time for them to start revising the chapters to score well. Get the MCQs for coordinate geometry to understand the concept well. Students are suggested to practise these questions by themselves and then verify the answers. Get important questions for class 10 Maths here as well.
Click here to download the PDF of additional MCQs for Practice on Coordinate Geometry, Chapter of Class 10 Maths along with answer key:
Download PDF
Students can also get access to Coordinate Geometry Class 10 Notes here.
MCQs for Coordinate Geometry
1. The points (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0) form a quadrilateral of type:
(a) Square
(b) Rectangle
(c) Parallelogram
(d) Rhombus
Answer: (a) Square
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:
Length of the side AB is:
\(AB= \sqrt{(1+1)^{2} + (0+2)^{2}}= 2\sqrt{2}\)Length of the side BC is:
\(BC= \sqrt{(-1-1)^{2} + (2-0)^{2}}= 2\sqrt{2}\)Length of the side CD is:
\(CD= \sqrt{(-3+1)^{2} + (0-2)^{2}}= 2\sqrt{2}\)Length of the side DA is:
\(DA= \sqrt{(-3+1)^{2} + (2)^{2}}= 2\sqrt{2}\)Let us calculate the diagonals.
\(AC= \sqrt{(-1+1)^{2} + (2+2)^{2}}= 4\)And
\(BD= \sqrt{(1+3)^{2} + (0-0)^{2}}= 4\)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
Answer: (a) 2
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)
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 AB.
4. The distance of point A(2, 4) from x-axis is
(a) 2 units
(b) 4 units
(c) -2 units
(d) -4 units
Answer: (b) 4 units
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
Answer: (b) 2√10
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
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.
7. The point 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)
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)
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
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.
9. The coordinates of a point P, where PQ is the diameter of a circle whose centre is (2, – 3) and Q is (1, 4) is:
(a) (3, -10)
(b) (2, -10)
(c) (-3, 10)
(d) (-2, 10)
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}By equating the corresponding coordinates,
(x+1)/2 = 2
x+1=4
x=3
And
(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.units
(b) 24 sq.units
(c) 30 sq.units
(d) 32 sq.units
Answer: (b) 24 sq.units
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.units.
11. The distance of the point P(–6, 8) from the origin is
(a) 8 units
(b) 2√7 units
(c) 10 units
(d) 6 units
Answer: (c) 10 units
Explanation:
We know that the distance of a point P(x, y) from the origin is √(x2 + y2).
Thus, the distance of the point P(-6, 8) from the origin = √[(-6)2 + (8)2]
= √(36 + 64)
= √100
= 10 units
12. The distance between the points (0, 5) and (–5, 0) is
(a) 5 units
(b) 5√2 units
(c) 2√5 units
(d) 10 units
Answer: (b) 5√2 units
Explanation:
Let the given points be:
A(0, 5) = (x1, y1)
B(-5, 0) = (x2, y2)
The distance between A and B = √[(x2 – x1)2 + (y2 – y1)2]
= √[(-5 – 0)2 + (0 – 5)2]
= √(25 + 25)
= √50
= √(2 × 25)
= 5√2
13. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(a) 5
(b) 12
(c) 11
(d) 7 + √5
Answer: (b) 12
Solution:
Let A(0, 4), B(0, 0), and C(3, 0) be the vertices of a triangle ABC.
Perimeter of the triangle = AB + BC + CA
Using the distance formula,
AB = √[(0)2 + (4)2] = √16 = 4
BC = √[(3)2 + (0)2] = √9 = 3
CA = √[(3)2 + (4)2] = √(9 + 16) = √25 = 5
Therefore, perimeter = 4 + 3 + 5 = 12
14. The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is
(A) (a + b + c)2
(B) 0
(C) a + b + c
(D) ab
Answer: (b) 0
Explanation:
Let the vertices of a triangle be:
A = (x1, y1) = (a, b + c)
B = (x2, y2) = (b, c + a)
C = (x3, y3) = (c, a + b)
Area of triangle ABC = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= (1/2)[a(c + a – a – b) + b(a + b – b – c) + c(b + c – c – a)]
= (1/2)[a(c – b) + b(a – c) + c(b – a)]
= (1/2)[ac – ab + ab – bc + bc – ac]
= (1/2)(0)
= 0
15. The point which lies on the perpendicular bisector of the line segment joining the points A(–2, –5) and B(2, 5) is
(a) (0, 0)
(b) (0, 2)
(c) (2, 0)
(d) (–2, 0)
Answer: (a) (0, 0)
Explanation:
We know that, the perpendicular bisector of any line segment divides the line segment into two equal parts.
That means, the perpendicular bisector of the line segment always passes through the midpoint of the line segment.
Thus, the midpoint of the line segment joining the points A(-2, -5) and B(2, 5) = [(-2 + 2)/2, (-5 + 5)/2]
= (0/2, 0/2)
= (0, 0)
16. If the points A(1, 2), O(0, 0) and C(a, b) are collinear, then
(a) a = b
(b) a = 2b
(c) 2a = b
(d) a = –b
Answer: (c) 2a = b
Explanation:
Let the given points be:
A(1, 2) = (x1, y1)
O(0, 0) = (x2, y2)
C(a, b) = (x3, y3)
If three points are collinear, then the area of the triangle formed by these points is equal to 0.
So, (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0
(1/2)[1(0 – b) + 0(b – 2) + a(2 – 0)] = 0
-b + 0 + 2a = 0
2a = b
17. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, then the value of p is
(a) 4
(b) -6
(c) 7
(d) -2
Answer: (c) 7
Explanation:
Given that A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram.
We know that diagonals of a parallelogram bisect each other.
So, the coordinates of the midpoint of AC = coordinates of the midpoint of BD
⇒ [(6 + 9)/2, (1 + 4)/2] = [(8 + p)/2, (2 + 3)/2]
⇒ (15/2, 5/2) = [(8 + p)/2, 5/2]
By equating the x-coordinates,
(8 + p)/2 = 15/2
⇒ 8 + p = 15
⇒ p = 7
18. A line intersects the y-axis and x-axis at the points P and Q, respectively. If (2, -5) is the midpoint of PQ, then the coordinates of P and Q are, respectively
(a) (0, -5) and (2, 0)
(b) (0, 10) and (-4, 0)
(c) (0, 4) and (-10, 0)
(d) (0, -10) and (4, 0)
Answer: (d) (0, -10) and (4, 0)
Explanation:
Let (0, y) and (x, 0) be the vertices of points P and Q respectively.
Given that (2, -5) is the midpoint of PQ.
So, [(0 + x)/2, (y + 0)/2] = (2, -5)
(x/2, y/2) = (2, -5)
Now, by equating the corresponding coordinates,
x/2 = 2, y/2 = -5
x = 4, y = -10
Hence, the coordinates of points P and Q are respectively (0, -10) and (4, 0).
19. The perpendicular bisector of the line segment joining the points A(1, 5) and B(4, 6) cuts the y-axis at
(a) (0, 13)
(b) (0, –13)
(c) (0, 12)
(d) (13, 0)
Answer: (a) (0, 13)
Explanation:
As we know, the perpendicular bisector of line segment AB, perpendicular at AB and passes through the mid – point of AB.
Let P be the mid – point of AB.
Mid – point of AB = [(1 + 4)/2, (5 + 6)/2]
P = (5/2, 11/2)
Now, the slope of the line AB = (6 – 5)/(4 – 1) = 1/3
So, the slope of the bisector = -1/slope of line AB = -1/(1/3) = -3
Equation of the line through the point P(5/2, 11/2) with slope -3 is given by using the point-slope form as:
y – (11/2) = -3[x – (5/2)]
(2y – 11)/2 = -3[(2x – 5)/2]
2y – 11 = -6x + 15
2y = -6x + 15 + 11
2y = -6x + 26
y = -3x + 13
This is of the form y = mx + c
Here, c = 13 is the y-intercept.
Therefore, the perpendicular bisector of the line segment joining the points A(1, 5) and B(4, 6) cuts the y-axis at (0, 13).
20. The fourth vertex D of a parallelogram ABCD whose three vertices are A(–2, 3), B(6, 7) and C(8, 3) is
(a) (0, 1)
(b) (0, –1)
(c) (–1, 0)
(d) (1, 0)
Answer: (b) (0, –1)
Explanation:
Given that A(–2, 3), B(6, 7) and C(8, 3) are three vertices of a parallelogram.
Let D(x, y) be the fourth vertex.
Assume that the vertices A, B, C and D are taken in order.
We know that diagonals of a parallelogram bisect each other.
So, the coordinates of the midpoint of AC = coordinates of the midpoint of BD
[(-2 + 8)/2, (3 + 3)/2] = [(6 + x)/2, (7 + y)/2]
(6/2, 6/2) = [(6 + x)/2, (7 + y)/2]
By equating the coordinates,
(6 + x)/2 = 6/2, (7 + y)/2 = 6/2
6 + x = 6, 7 + y = 6
x = 0, y = -1
Therefore, the coordinates of the fourth vertex D = (0, -1)