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cbse 10th Standard Maths Coordinate Geometry Important Questions And Answers Study Material - QB365 Set B

Published on: 22/10/2025

Question Set B

cbse 10th Standard Maths Coordinate Geometry Important Questions And Answers Study Material - QB365

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Std: 10th Standard CBSE - Maths

chapter: Coordinate Geometry - Important Questions And Answers

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Questions:

Generic 5

1 x 5 = 5

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer. (4,5), (7,6), (4,3), (1,2)

Generic 2

4 x 2 = 8

Show that the points (a,a), (-a, -a) and \(\left( -\sqrt { 3 } a,\sqrt {3 } a \right) \) are the vertices of an equidistant triangle.

If P(2,1), Q(4,2), R(5,4) and S(3,3) are vertices of a quadrilateral, find the area of the quadrilateral PQRS.

If the points (10,5),(8,4) and (6,6) are the mid-points of the sides of a triangle, find its vertices.

Find the relation between x and y such that the point (x,y) is equidistant from the points (7,1) and (3,5).

Generic 3

3 x 3 = 9

Find the coordinates of the point which divies the line segment joining the points (3,4) and (-5, -7) internally in the following ratios 2 : 3.

If A(4, -1), B(5, 3), C(2, y) and D(1, 1) are the vertices a parallelogram ABCD, find y.

Find the ratio in which the line segment joining the points A(3, -3) and B(-2, 7) is divided by x-axis. Also find the co-ordinate of point of division.

Multiple Choice Question

5 x 1 = 5

Find the ratio in which the line segment joining (2,-3) to and (5, 6) is divided by X-axis.

(a)

1 : 2

(b)

2 : 1

(c)

2 : 5

(d)

5 : 2

10.

The perpendicular bisector of a line segment A(-8,0) and B(8,0) passes through a point (0, k). The value of k is

(a)

0 only

(b)

0 or 8 only

(c)

any real number

(d)

any non-zero real number

11.

The condition that the point (x,y) may lie on the line joining (3,4) and (-5,-6) is

(a)

-5x+4y+1=0

(b)

-5x-4y+1=0

(c)

5x+4y+1=0

(d)

5x-4y+1=0

12.

The perimeter of a triangle with vertices (0, 4) (0, 0) and (3, 0) is

(a)

(b)

(c)

(d)

13.

Find the coordinates of the point equidistant from the points A(5, 1), B(–3, –7) and C(7, –1)

(a)

(2, –4)

(b)

(3, –6)

(c)

(4, 7)

(d)

(8, –6)

Assertion and reason

1 x 1 = 1

14.

Assertion: Three points A, B and C are such that AB + BC > AC, then they are collinear.
Reason: Three points are collinear if they lie on a straight line.
Codes :
(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
(c) If Assertion is correct but Reason is incorrect.
(d) If Assertion is incorrect but Reason is correct.

Case Study Questions

1 x 4 = 4

15.

Aditya asked carpenter to make front door of his guest house. The carpenter suggested him a design which is plotted on a graph as shown in below figure:

(i) What is the length of the line segment AB?

(a) \(\sqrt{10}\) units

(b) \(\sqrt{11}\) units

(c) \(\sqrt{13}\) units

(d) \(\sqrt{14}\) units

(ii) Is AB = AC?

(a) yes

(b) no

(c) can't be determined

(d) none of these

(iii) The coordinates of the midpoint of BE are

(a) (4,-3)

(b) (-3,4)

(c) (-3,-4)

(d) (-4,4)

(iv) Midpoint of ED will lie on:

(a) x-axis

(b) y-axis

(c) x = y

(d) x = 2y

(v) If we join BD, then the y-axis divides BD in the ratio:

(a) 1:1

(b) 2:1

(c) 1:2

(d) 2:3

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Answers Key:

Generic 5

1 x 5 = 5

Let points be A(4,5), B(7,6), C(4,3) and D(1,2)
\(AB=\sqrt { \left( 7-4 \right) ^{ 2 }+\left( 6-5 \right) ^{ 2 } } =\sqrt { 9+1 } =\sqrt { 10 } \)
\(BC=\sqrt { \left( 4-7 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 } } =\sqrt { 9+9 } =3\sqrt { 2 } \)
\(CD=\sqrt { \left( 1-4 \right) ^{ 2 }+\left( 2-3 \right) ^{ 2 } } =\sqrt { 9+1 } =\sqrt { 10 } \)
\(AD=\sqrt { \left( 1-4 \right) ^{ 2 }+\left( 2-5 \right) ^{ 2 } } =\sqrt { 9+9 } =3\sqrt { 2 } \)
\(AC=\sqrt { \left( 4-4 \right) ^{ 2 }+\left( 3-5 \right) ^{ 2 } } =\sqrt { 4 } =2\)
\(BD=\sqrt { \left( 1-7 \right) ^{ 2 }+\left( 2-6 \right) ^{ 2 } } =\sqrt { 36+16 } =\sqrt { 52 } =2\sqrt { 13 } \)
Here, AB=CD=, BC=AD and AC≠BD
∴ The quadrilateral ABCD is a parallelogram.

Generic 2

4 x 2 = 8

Let A ( a,a ), B(-a, -a) and C\(\left( -\sqrt { 3 } a,\sqrt {3 } a \right) \)
Here \(AB=\sqrt { { \left( a+a \right) }^{ 2 }+{ \left( a+a \right) }^{ 2 } } =2\sqrt { 2 } a\)
\(BC=\sqrt { { \left( -a+\sqrt { 3 } a \right) }^{ 2 }+{ \left( -a-\sqrt { 3 } a \right) }^{ 2 } } =2\sqrt { 2 } a\)
\(CA=\sqrt { { \left( a+\sqrt { 3 } a \right) }^{ 2 }+{ \left( a-\sqrt { 3 } a \right) }^{ 2 } } =2\sqrt { 2 } a\)
Since AB = BC = AC, therefore ABC is an equilateral triangle.

Area of △PQR = \(\frac{1}{2}\)[2(2-4)+4(4-1)+5(1-2)]
=\(\frac{1}{2}\)[2x-2+4x3+5x-1]=\(\frac{1}{2}\)[-4+12-5]=\(\frac{3}{2}\)sq.unit.
Area of △PRS =\(\frac{1}{2}\)[2(4-3)+5(3-1)+3(1-4)| = \(\frac{1}{2}\)|2+10-9|=\(\frac{3}{2}\)sq.unit.
∴ Area of quadrilateral = Area of △PQR+ Area of △PRS = \(\frac{3}{2}\)+\(\frac{3}{2}\)=3 sq.units.

Let A(x₁,y₂), B (x₂, y₂) and c(x₃,y₃) be the vertices of a triangle D (10, 5), E (8, 4) and F(6,6) are mid-points of sides BC, CA and AB respectively.
Therefore \(\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) =\left( 6,6 \right) \)
\(\Rightarrow { x }_{ 1 }+{ x }_{ 2 }=12\) ...(i)
\({ y }_{ 1 }+y_{ 2 }=12\) ...(ii)
\( \left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) =\left( 10,5 \right) \)
\({ x }_{ 2 }+x_{ 3 }=20\) ...(iii)
\(\ { y }_{ 2 }+y_{ 3 }=10\) ...(iv)
\(\ \left( \frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } \right) =\left( 8,4 \right)\)
\(\Rightarrow \ { x }_{ 1 }+x_{ 3 }=16\)
\(\ { y }_{ 1 }+y_{ 3 }=8\) ...(v)
Adding (i), (iii) and (v) we get,
\(2({ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 })=48\) ...(vi)
\(\Rightarrow { x }_{ 1 }+{ x }_{ 2 }+x_{ 3 }=24\)
Adding (ii), (iv) and (vi) we get,
\(2({ y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 })=15\) ...(vii)
From (ii), (iv), (vi) and (xi) we get,
\({ y }_{ 1 }=5,\ { y }_{ 2 }=7,\ { y }_{ 3 }=3\) ...(viii)
Coordinates of vertices are A(4,5) B(8,7) and C(12,3).

Since the point P(x,y) is equidistant from the points A(7, 1) and B(3, 5).
Therefore PA = PB ...(i)
Using distance formula
\(PA=\sqrt { \left( x-7 \right) ^{ 2 }+\left( y-1 \right) ^{ 2 } } \)
\(=\sqrt { { x }^{ 2 }+{ 7 }^{ 2 }-2.7x+{ y }^{ 2 }+{ 1 }^{ 2 }-2.y.1 } =\sqrt { { x }^{ 2 }+{ y }^{ 2 }+49+1-14x-2y } \)
\(PB=\sqrt { \left( x-3 \right) ^{ 2 }+\left( y-5 \right) ^{ 2 } } \) (using distance formula)
\(=\sqrt { { x }^{ 2 }+{ 3 }^{ 2 }-2.3x+{ y }^{ 2 }+{ 5 }^{ 2 }-2.5.y } =\sqrt { { x }^{ 2 }+9-6x+{ y }^{ 2 }+25-10y } \)
Substituting the values of PA and PB in (i). we get
\(\sqrt { { x }^{ 2 }+{ y }^{ 2 }+50-14x-2y } =\sqrt { { x }^{ 2 }+{ y }^{ 2 }-6x-10y+34 } \)
Squaring both sides, We get
\({ x }^{ 2 }+{ y }^{ 2 }+50-14x-2y={ x }^{ 2 }+{ y }^{ 2 }-6x-10y+34\)
⇒ 50-34=14x+2y-6x-10y ⇒16=8x-8y ⇒x-y=2

Generic 3

3 x 3 = 9

\(\left(\frac{-1}{5}, \frac{-2}{5}\right)\)

Diagonals of a parallelogram bisect each other.
Mid-points of AC and BD are same
\(\Rightarrow \quad \left( 3,\frac { -1+y }{ 2 } \right) =\left( 3,2 \right) \)
\(\frac { -1+y }{ 2 } =2\quad \Rightarrow \quad y=5\)

Using section formula, we get,
\(0=\frac { 1\left( -3 \right) +k\left( 7 \right) }{ 1+k } \)
Let the ratio be k : 1,
\(\Rightarrow \ k=\frac{3}{7}\)
\(\Rightarrow\) Ratio = 3 : 7
Also, \(x=\frac { { m }_{ 2 }{ x }_{ 1 }+{ m }_{ 1 }{ x }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
\(=\frac { 1\left( 3 \right) +k\left( -2 \right) }{ 1+k } \)
\(x=\frac { 3-2\times \frac { 3 }{ 7 } }{ 1+\frac { 3 }{ 7 } } =\frac { 3 }{ 2 } \)
\(\therefore\) Co-ordinates of point are \(\left( \frac { 3 }{ 2 } ,0 \right) \)

Multiple Choice Question

5 x 1 = 5

(a)
1 : 2

10.

11.

(d)
5x-4y+1=0

12.

(b)
12

13.

(a)
(2, –4)

Assertion and reason

1 x 1 = 1

14.

(d) Assertion Three points A, B, C are collinear if and only if AB + BC = AC, but AB + BC > AC.
. Ii A, B, C att: not collinear.

Case Study Questions

1 x 4 = 4

15.

(i) (c): \(\mathrm{AB}=\sqrt{(0+3)^{2}+(8-6)^{2}}\)
\(=\sqrt{13} \text { units }\)
(ii) (a): \(A C=\sqrt{(0-3)^{2}+(8-6)^{2}}\)
\(=\sqrt{13} \text { units }=\mathrm{AB}\)
(iii) (b): coordinates of the midpoint of BE are \(\left(\frac{-3-3}{2}, \frac{6+2}{2}\right)=(-3,4)\)
(iv) (b): y-axis
(v) (a): 1:1