MCQ Questions for Class 10 Maths Chapter 7 Coordinate Geometry

Coordinate Geometry Class 10 MCQ Questions with Answers

Question 1.

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

(A) 2
(B) 3
(C) 1
(D) 5
Answer:
(B) 3

Explanation:
Since coordinate of a given point is the distance of point from x-axis.

Question 2.

The distance between the points A (0, 6) and B (0,-2) is:

(A) 6
(B) 8
(C) 4
(D) 2
Answer:
(B) 8

Explanation:
The distance between two points (x₁, y₁) and (x₂, y₂ ) is given as,
d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
Where, x1, = 0, y1 = 6 and x2 = 0, y2 = -2 So, distance between A (0, 6) and B (0, -2):
AB = \(\sqrt{(0-0)^{2}+(-2-6)^{2}}\)
= \(\sqrt{0+(-8)^{2}}/latex]
= [latex]\sqrt{8^{2}}\)
= 8

Question 3.

The distance of the point P (-6,8) from the origin is:

(A) 8
(B) 2√7
(C) 10
(D) 6
Answer:
(C) 10

Explanation:
Distance between two points (x₁, y₁) and (x₂, y₂) is given as,
d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
Where, x₁= -6, y₁ =8 and x₂ = 0, y₂ = 0 So, distance between P (-6,8) and origin O (0,0) is given by,
PO = \(\sqrt{[0-(-6)]^{2}+[(0-8)]^{2}}\)
= \(\sqrt{(6)^{2}+(-8)^{2}}\)
= \(\sqrt{36+64}\)
= \(\sqrt{100}\)
= 10

Question 4.

The distance between the points (0, 5) and (-5, 0) is:

(A) 5
(B) 5√2
(C) 2√5
(D) 10
Answer:
(B) 5√2

Explanation:
Distance between two points (x₁,y₁) and (x₂,y₂) is given as,
= \(\sqrt{(-5-0)^{2}+(0-5)^{2}} \)
Where, x₁= 0, y₁= 5 and x₂ = -5, y₂ = 0 So that, distance between the point (0, 5) and I (-5,0),
= \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{(-5-0)^{2}+(0-5)^{2}}\)
= \(\sqrt{25+25}\)
= \(\sqrt{50}\)
= 5√2

Question 5.

AOBC is a rectangle whose three vertices are vertices A (0, 3), O (0, 0) and B (5, 0). The length of its diagonal is:

(A) 5
(B) 3
(C) √34
(D) 4
Answer:
(C) √34

Explanation:
According to the question, a triangle can be represented as:

∴ Distance between the points A (0,3) and B (5, 0) is
AB = \(\sqrt{(5-0)^{2}+(0-3)^{2}}\)
= \(\sqrt{25+9}\)
= \(\sqrt{34}\)
Hence, the required length of diagonal is √34.

Question 6.

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
Explanation:
Since, Perimeter of triangle = OA + AB + OB

Here OA = 4 units, AB = 5 units (using pythagoras theorem in triangle AOB) and OB =3 units.
Therefore, Perimeter of triangle = 4 + 5+3
= 12 units.
Question 7.

The points (-4,0), (4,0) and (0,3) are the vertices of a:

(A) right triangle
(B) isosceles triangle
(C) equilateral triangle
(D) scalene triangle E
Answer:
(B) isosceles triangle

Explanation:
Let X (-4, 0), Y (4, 0) and Z (0, 3) are the given vertices.
Now, distance between X ( -4,0) and Y (4, 0),
XY = \(\sqrt{[4-(-4)]^{2}+(0-0)^{2}}\)
= \(\sqrt{(4+4)^{2}}\)
= \(\sqrt{8^{2}}\)
= 8
Distance between Y(4, 0) and Z (0, 3),
YZ = \(\sqrt{(0-4)^{2}+(3-0)^{2}}\)
= \(\sqrt{16+9}\)
= \(\sqrt{25}\)
= 5
Distance between Y (4,0) and Z (0, 3),
ZX = \(\sqrt{[0-(-4)]^{2}+(3-0)^{2}}\)
= \(\sqrt{16+9}\)
= \(\sqrt{25}\)
= 5
As, YZ = ZX, i.c., two sidps of the triangle ane H equal.
So that, the AXYZ is an isosceles triangle.

Question 8.

The coordinates of the point which is equidistant from the three vertices of the ∆AOB as shown in the figure is:

(A) (x, y)
(B) (y,x)
(C) \(\left(\frac{x}{2}, \frac{y}{2}\right)\)
(D) \(\left(\frac{y}{2}, \frac{x}{2}\right)\)
Answer:
(A) (x, y)

Explanation:
From the given figure, coordinates of three vertices of a triangle arc: O (0, 0), A (0,2y) and B (2x, 0). Suppose the required point be P whose coordinates are (h, k) Now, P is equidistant trom the three vertices of ∆AOB, therefore,
PO = PA = PB
or (PO)² = (PA)² = (PB)²
By using distance formula, we get:
\(\left(\sqrt{(h-0)^{2}+(k-0)^{2}}\right)^{2}\) = \(\left(\sqrt{(h-0)^{2}+(k-2 y)^{2}}\right)^{2}\)
= \(\left(\sqrt{(h-2 x)^{2}+(5-0)^{2}}\right)^{2}\)
⇒ h² + k² = h² +(k – 2y)2 = (h – 2x)² + k² ………(i)
⇒ k² = k² + 4y² – 4yk
⇒ 0 = 4y² – 4yk
⇒ 4y (y – k) = 0
⇒ y = k
Taking first and second parts of EQuestion (i),
h² + k² = (h- 2x)² + k²
⇒ h² = h² + 4x² – 4xh
⇒ 4x (x – h) = 0
⇒ h = x
Required coordinates of the point are (h, k) = (x, y)

Assertion and Reason Based MCQs

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True

Question 1.

Assertion (A): If the distance between the point (4, p) and (1,0) is 5, then the value of p is 4.
Reason (R): The point which divides the line segment joining the points (7, – 6) and (3,4) in ratio 1: 2 internally lies in the fourth quadrant.

Answer:
(D) A is false and R is True

Explanation:
In case of assertion:
Distance between two points (x₁,y₁) and (x², y²) is given as,
d = \(\sqrt{\left(x_{2}+x_{1}\right)^{2}+\left(y_{2}+y_{1}\right)^{2}}\)
where,
(x₁,y₁) = (4, p)
(x₂,y₂) = (1, 0)
And, d = 5
Put the values, we have
5² = (1 — 4)² + (0 – p)²
25 = (-3)² + (-p)²
25-9 = p²
16 = p²
+4, -4 = p
∴ Assertion is incorrect.
In case of reason:
Let (x, y) be the point
Then, x = \(\frac{m x_{2}+n x_{1}}{m+n}\) and y = \(\frac{m y_{2}+n y_{1}}{m+n}\) ………..(i)
Here, x₁ = 7, y₁= -6, x₁ = 3, y₁ = 4, m = 1 and n = 2
∴ x = \(\frac{1(3)+2(7)}{1+2}\)
⇒ x = \(\frac{3+14}{3}=\frac{17}{3}\)
And, y = \(\frac{1(4)+2(-6)}{1+2}\)
⇒ y = \(\frac{4-12}{3}=-\frac{8}{3}\)
So, the required point (x,y) = \(\left(\frac{17}{3},-\frac{8}{3}\right)\) lies in IV ^th quadrant.
∴ Reason is correct.
Hence, assertion is incorrect but reason is correct.

Question 2.

Assertion (A): AABC with vertices A(-2, 0), B(2, 0) and C(0,2) is similar to ADEF with vertices D(-4,0), E(4,0) and F(0,4).
Reason (R): A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6,8) lies outside the circle.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion:
By using Distance Formula,
∴Distance between A (-2,0) and B (2,0),
AB = \(\sqrt{[2-(-2)]^{2}+(0-0)^{2}}\)=4
[∴distance between the points (x1, y1) and (x2,y2)
D = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
Similarly, distance between B (2,0) and C (0, 2),
BC = \(\sqrt{(0-2)^{2}+(2-0)^{2}}\) = \(\sqrt{4+4}\) = 2√2
In ∆ABC, distance between C (0, 2) and A (-2,0),
CA = \(\sqrt{\left[0-(-2)^{2}\right]+(2-0)^{2}}\) = \(\sqrt{4+4}\) = 2√2
∴Distance between F (0,4) and D (-4, 0),
FD = \(\sqrt{(0+4)^{2}+(0-4)^{2}}\) = \(\sqrt{4^{2}+(4)^{2}}\) = 4√2
∴Distance between F (o,4) and E (4,0) and D (-4, 0),
FE = \(\sqrt{(0-4)^{2}+(4-0)^{2}}\) = \(\sqrt{4^{2}+4^{2}}\) = 4√2
and distance between E (4,0) and D (-4, 0),
ED = \(\sqrt{[4-(-4)]^{2}+(0)^{2}}\) = \(\sqrt{8^{2}}\) = 4√2
Now, \(\frac{A B}{O E}\) = \(\frac{4}{9}\) = \(\frac{1}{2}\),
\(\frac{A C}{D E}\) = \(\frac{2 \sqrt{2}}{4 \sqrt{2}}\) = \(\frac{1}{2}\)
\(\frac{B C}{E F}\) = \(\frac{2 \sqrt{2}}{4 \sqrt{2}}\)= \(\frac{1}{2}\)
∴ \(\frac{A B}{D E}\) = \(\frac{A C}{D F}\) = \(\frac{B C}{E F}\)
Here, we see that sides of AABC and AFDE are proportional.
Therefore, by SSS similarity rule, both the triangles are similar.
∴ Assertion is correct.
In case of reason:
Faint Q (6, 8) will lie outside the circle if its distance from the centre of circle is greater than the radius of the circle.
Distance between centre O (0,0) and P (5,0),
OP = \(\sqrt{(5-0)^{2}+(0-0)^{2}}\)
= \(\sqrt{5^{2}+0^{2}}\)
= 5
As, point P lies at the circle, therefore, OP = Radius of circle.
Distance between centre O (0,0) and Q (6,8),
OQ = \(\sqrt{(6-0)^{2}+(8-0)^{2}}\)
= \(\sqrt{6^{2}+8^{2}}\)
= \(\sqrt{36+64}\)
= \(\sqrt{100}\)
= 10
As OQ > OP, therefore, point Q (6, 8) lies outside of the circle.
∴ Reason is correct.
Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Question 3.

Assertion (A): The ordinate of a point A on y-axis is 5 and B has coordinates (-3,1). Then the length of AB is 5 units.
Reason (R): The point A(2, 7) lies on the perpendicular bisector of line segment joining the points P(6,5) and Q(0, -4).

Answer:
(C) A is true but R is false

Explanation:
In case of assertion:
Here, A → (0,5) and B → (-3,1)
AB = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{(-3-0)^{2}+(1-5)^{2}}\)
= \(\sqrt{9+16}\)
= \(\sqrt{25}\)
= 5 units
∴ Assertion is correct.
In case of reason:
If A (2, 7) lies on perpendicular bisector of P (6,5) and Q (0, -4), then
AP = AQ
∴ By using Distance Formula,
AP = \(\sqrt{(6-2)^{2}+(5-7)^{2}}\)
= \(\sqrt{(4)^{2}+(-2)^{2}}\)
= \(\sqrt{20}\)
And AQ = \(\sqrt{(0-2)^{2}+(-4-7)^{2}}\)
= \(\sqrt{(-2)^{2}+(-11)^{2}}\)
= \(\sqrt{125}\)
As, AP AQ
Thercfore, A does not lie on the perpendicular bisector of PQuestion
∴ Reason is incorrect.
Hence, assertion is correct but reason is incorrect.

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following text and answer the below questions:
The diagram show the plans for a sun room. It will be built onto the wall of a house. The four walls of the sun room are square clear glass panels. The roof is made using,

• • Four clear glass panels, trapezium in shape, all of the same size
  • One tinted glass panel, half a regular octagon in shape

Question 1.

Refer to Top View, find the mid-point of the seg¬ment joining the points J(6,17) and 1(9,16).

(A) \(\frac{33}{2}, \frac{15}{2}\)
(B) \(\frac{3}{2}, \frac{1}{2}\)
(C) \(\frac{15}{2}, \frac{33}{2}\)
(D) \(\frac{1}{2}, \frac{3}{2}\)
Answer:
(C) \(\frac{15}{2}, \frac{33}{2}\)

Explanation:
Mid-point of J(6,17) and 1(9,16) is
x = \(\frac{6+9}{2}\) and y = \(\frac{17+16}{2}\)
x = \(\frac{15}{2}\) and y = \(\frac{33}{2}\)

Question 2.

Refer to front View, the distance of the point P from the y-axis is:

(A) 4
(B) 15
(C) 19
(D) 25
Answer:
(A) 4

Explanation:
The distance of the point P from the Y-axis = 4.

Question 3.

Refer to front view, the distance between the points A and S is

(A) 4
(B) 8
(C) 14
(D) 20
Answer:
(C) 14

Explanation:
A’s coordinates = (1,8)
S’s coordinates = (15,8)
Then, AS = \(\left|\sqrt{(15-1)^{2}+(8-8)^{2}}\right|\)
= \(\sqrt{(14)^{2}}\)
= 14.

Question 4.

Refer to front view, find the co-ordinates of the point which divides the line segment joining the points A and B in the ratio 1: 3 internally.

(A) (8.5,2.0)
(B) (2.0,9.5)
(C) (3.0,7.5)
(D) (2.0,8.5)
Answer:
(D) (2.0,8.5)

Explanation:
The coordinates of A = (1,8) The coordinates of B = (4,10) Also, m = 1 and n = 3
Then, (x,y) = \(\left(\frac{1 \times 4+3 \times 1}{1+3}, \frac{1 \times 10+3 \times 8}{1+3}\right)\)
= \(\left(\frac{7}{4}, \frac{34}{4}\right)\)
= (1. 75,8.5)

Question 5.

Refer to front view, if a point (x, y) is equidistant from the Q(9,8) and S(17,8), then

(A) x + y = 13
(B) x – 13 = 0
(C) y – 13 = 0
(D) Y – y = 13
Answer:
Option (B) is correct

Explanation:
Let point be P(x, y)
PQ² = PS²
or, (x – 9)² + (y – 8)² = (x – 17)² + (y – 8)² or, x – 13 =0

II. Read the following text and answer the below questions:
Ayush Starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office.
(Assume that all distances covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in km.

Question 1.

What is the distance between house and bank?

(A) 5 km
(B) 10 km
(C) 12 km
(D) 27 km
Answer:
(A) 5 km

Explanation:
Since,
Distance between two points (x₁, y₁) and (x₂, y₂).
d = \(\left|\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right|\)
Now, distance between house and bank,
= \(\left|\sqrt{(5-2)^{2}+(8-4)^{2}}\right|\)
= \(\left|\sqrt{(3)^{2}+(4)^{2}}\right|\)
= \(|\sqrt{9+16}|\)
= \(\sqrt{25}\)
= 5 km

Question 2.

What is the distance between Daughter’s School and bank ?

(A) 5 km
(B) 10 km
(C) 12 km
(D) 27 km
Answer:
(B) 10 km

Explanation:
Distance between bank and daughter’s school,
= \(\left|\sqrt{(13-5)^{2}+(14-8)^{2}}\right|\)
= \(\left|\sqrt{(8)^{2}+(6)^{2}}\right|\)
= \(|\sqrt{64+36}|\)
= \(|\sqrt{100}|\)
= 10 km

Question 3.

What is the distance between house and office?

(A) 24.6 km
(B) 26.4 km
(C) 24 km
(D) 26 km
Answer:
(B) 26.4 km

Explanation:
Distance between house to office,
= \(\vec{a}\)
= \(\vec{a}\)
= \(\vec{a}\)
= \(\vec{a}\)
= 24. 59
= 24.6 km

Question 4.

What is the total distance travelled by Ayush to reach the office?

(A) 5 km
(B) 10 km
(C) 12 km
(D) 27 km
Answer:
(D) 27 km

Explanation:
Distance between daughter’s school and office.
= \(\left|\sqrt{(13-13)^{2}+(26-14)^{2}}\right|\)
= \(\left|\sqrt{0+(12)^{2}}\right|\)
= 12 km
Total distance (House + Bank + School + Office) travelled = 5 + 10 + 12 = 27 km

Question 5.

What is the extra distance travelled by Ayush?

(A) 2 km
(B) 2.2 km
(C) 2.4 km
(D) none of these
Answer:
(C) 2.4 km

Explanation:
Extra distance travelled by Ayush in reaching his office = 27 – 24.6 = 2.4 km.

III. Read the following text and answer the below questions:
In order to conduct Sports Day activities in your School, lines have been drawn with chalk powder at a distance of 1 m each, in a rectangular shaped ground ABCD, 100 flowerpots have been placed at a distance of 1 m from each other along AD, as shown in given figure below. Niharika runs 1/4^th the distance AD on the 2nd line and posts a green (G) flag runs 1/5^th distance AD on the eighth line and posts a red (R) flag.

Question 1.

Find the position of green flag

(A) (2,25)
(B) (2,0.25)
(C) (25,2)
(D) (0,-25)
Answer:
Option (A) is correct.

Question 2.

Find the position of red flag

(A) (8,0)
(B) (20,8)
(C) (8,20)
(D) (8,0.2)
Answer:
(C) (8,20)

Question 3.

What is the distance between both the flags?

(A) √41
(B) √11
(C) √61
(D) √51
Answer:
(C) √61

Explanation:
Position of Green flag = (2,25) Position of Red flag = (8,20)
Distance between both the flags
\(\sqrt{(8-2)^{2}+(20-25)^{2}}\) = \(\sqrt{6^{2}+(-5)^{2}}\)
= \(\sqrt{36+25}\)
= √61

Question 4.

If Rashini has to post a blue llag exactly halfway between the line segment joining the two flags, where should she post her flag?

(A) (5,22.5)
(B) (10,22)
(C) (2,8.5)
(D) (2.5,20)
Answer:
(A) (5,22.5)

Explanation:
Position of blue flag Mid-point of line segment joining the green and red flags
= \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)\)
= (5, 22.5)

Question 5.

If Joy has to post a flag at one-fourth distance from green flag, in the line segment joining the green and red flags, then where should he post his flag ?

(A) (3.5,24)
(B) (0.5,12.5)
(C) (2.25,8.5)
(D) (25,20)
Answer:
(A) (3.5,24)

Explanation:
Position of Joy’s flag = Mid-point of line segment joining green and blue flags
= \(\left[\frac{2+5}{2}, \frac{25+22.5}{2}\right]\)
= [3.5,23.75] ~ [3.5,24]

IV. Read the following text and answer the below questions:
The class X students school in krishnagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

Question 1.

Taking A as origin, find the coordinates of P.

(A) (4,6)
(B) (6,4)
(C) (0,6)
(D) (4,0)
Answer:
(A) (4,6)

Question 2.

What will be the coordinates of R, if C is the origin?

(A) (8,6)
(B) (3,10)
(C) (10,3)
(D) (0,6)
Answer:
(C) (10,3)

Question 3.

What will be the coordinates of Q, if C is the origin?

(A) (6,13)
(B) (-6,13)
(C) (-13,6)
(D) (13,6)
Answer:
(D) (13,6)

Question 4.

Calculate the area of the triangles if A is the origin.

(A) 4.5
(B) 6
(C) 8
(D) 6.25
Answer:
(A) 4.5

Explanation:
Coordinates of P = (4,6)
Coordinates of Q = (3,2)
Coordinates of R = (6,5)
Area of triangle PQR = [ x₁ (y₂ – y₂ + x₂ (y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [4(-3) + 3(-1) + 6(4)]
= \(\frac{1}{2}\) [- 12 + (-3) + 24]
= \(\frac{1}{2}\)[- 12 + 21]
= \(\frac{1}{2}\)[9]
= 4.5 sQuestion units.

Question 5.

Calculate the area of the triangle if C is the origin.

(A) 8
(B) 5
(C) 6.25
(D) 4.5
Answer:
(D) 4.5

MCQ Questions for Class 10 Maths with Answers