Coordinate Geometry Chapter 7: NCERT Exemplar for Class 10

The NCERT Exemplar Class 10 Math Chapter 7 Coordinate Geometry is provided here in PDF format free for students to prepare for their CBSE exams. These exemplar problems and solutions are designed by subject matter experts with respect to the upgraded CBSE Syllabus(2024-25). The exemplar can further be used as an effective study tool and will help students to develop a better understanding about geometry in math and around the world. It also helps to tackle even the difficult questions that can be asked in the CBSE first and second-term exams.

In Chapter 7 of NCERT Exemplar Class 10 Solutions, students will be introduced to different formulas like distance and section formulas and they will need to understand their application properly. The chapter also deals with topics like the area of a triangle, and students need to work on problems based on them. This is the main objective of the chapter, which basically helps students focus more on applying the concepts of coordinate geometry to real-world scenarios.

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Question 1. The distance of point P (2, 3) from the x-axis is

(A) 2

(B) 3

(C) 1

(D) 5

Answer 1:(B) 3

Explanation: We already know that

(x, y) is the point on the Cartesian plane in the first quadrant.

Now,

X = Perpendicular distance from the Y – axis and

Y = Perpendicular distance from the X – axis

Thus, the perpendicular distance from the X-axis = y coordinate = 3

Source: Internet

 

Question 2. The distance between the points A (0, 6) and B (0, –2) will be

(A) 6

(B) 8

(C) 4

(D) 2

Answer 2: (B) 8

Explanation: For the distance formula:

D2 = (x2 – x1)2 + (y2 – y1)2

As per the question,

We get,

V1 = 0, v2 = 0

z1 = 6, z2 = – 2

For the distance formula,

D2 = (0 – 0)2 + ( – 2 – 6)2

D= √((0)2+ (-8)2)

D = √64

D = 8 units

Thus, the distance between A (0, 6) and B (0, 2) is 8

 

Question 3. The distance of the point P (–6, 8) from the origin will be

(A) 6

(B) 2√7

(C) 10

(D) 8

Answer 3: (C) 10 units

Explanation: For the distance formula:

D2 = (x2 – x1)2 + (y2 – y1)2

As per the question,

We get;

v1 = – 6, v2 = 0

u1 = 8, u2 = 0

D2 = [0 – ( – 6)]2 + [0 – 8]2

D= √((0-(-6))2+ (0-8)2

D= √((6)2+ (-8)2)

D = √(36 + 64)

D = √100

D = 10

Thus, the distance between P ( – 6, 8) and origin O (0, 0) is 10

 

Question 4. The distance between the points (0, 5) and (–5, 0) will be

(A) 10

(B) 5√2

(C) 2√5

(D) 5

Answer 4: (B) 5√ 2 units

Explanation: For the distance formula:

D2 = (x2 – x1)2 + (y2 – y1)2

As per the question,

We have now;

Z1 = 0, z2 = – 5

W1 = 5, w2 = 0

D2 = (( – 5) – 0)2 + (0 – 5)2

D= √(-5-0)2+ (0-5)2

D= √((-5)2+ (-5)2)

D = √(25 + 25)

D= √50= 5√2

Hence, the distance between (0, 5) and ( – 5, 0) = 5√ 2

 

Question 5. AOBC is the rectangle whose three vertices are vertices A (0, 3), O (0, 0), and B (5, 0). The length of the diagonal will be

(A) 5

(B) 3

(C) √34

(D) 4

Answer 5: (C) √34

Explanation: The three vertices are as follows: A = (0, 3), O = (0, 0), B = (5, 0)

We already know that the diagonals of the rectangle are of equal length,

Length of the diagonal AB = Distance between the points A and B

For the distance formula:

D2 = (x2 – x1)2 + (y2 – y1)2

As per the question,

We get;

Source: Internet

P1 = 0, p2 = 5

Q1 = 3, q2 = 0

D2 = (5 – 0)2 + (0 – 3)2

D= √((5-0)2 + (0-3)2)

D = √(25 + 9)= √34

Distance between the side A (0, 3) and B (5, 0) is √34

Thus, the length of the diagonal is √34

 

Question 6. Points P (3, 1), Q (12, –2), as well as R (0, 2), cannot be the vertices of the triangle.

Answer 6: The given statement is true.

Explanation:

Coordinates of the point P = (s1, t1) = (3, 1)

Coordinates of the point Q = (s2, t2) = (12, – 2)

Coordinates of the point R = (s3, t3) = (0, 2)

Area of the given ∆PQR =

∆ = 1/2 [ a2 (b3 – b1 ) + a1 (b2 – b3) + a3 (b1 – b2 )]

Δ = ½ [12(2 – 1) + 3 (–2 – 2) + 0{1 – (- 2)}]

Δ = ½ [12(1) + 3(- 4) + 0]

Δ = ½ (- 12 + 12) =0

Area of the ΔPQR = 0

As the points P(3, 1), Q (12, – 2) and R(0, 2) are collinear.

Thus, the points P(3, 1), Q(12, – 2) and R(0, 2) can’t be the vertices of the triangle.

 

Question 7. Name the type of the triangle formed by the points S (–5, 6), T (–4, –2), and U (7, 5).

Answer 7: The points are given as S (–5, 6), T (–4, –2), and U (7, 5)

By using the distance formula,

D = √ ((x2 – x1)2 + (y2 – y1)2)

ST = √((-4+5)² + (-2-6)²)

= √1+64

=√65

TU =√((7+4)² + (5+2)²)

=√121 + 49

=√170

SU =√((7+5)² + (5-6)²)

=√144 + 1

=√145

As all the sides are of different lengths, STU is the scalene triangle.

 

Question 8. Find out the points on the x-axis which are at the distance of 2√5 from the point (7, –4). How many of these points are there?

Answer 8: Assume the coordinates of the point= (x, 0) (given that the point lies on the x axis)

C1=7. c1=-4

E2=x. e2=0

Distance =√(x2-x1)2+ (y2-y1)2

As per the question,

2√5=√(x-7)2+ (0-(-4))2

On squaring both the L.H.S and R.H.S,

20=x2+49-14x+16

20=x2+65-14x

0=x2-14x+45

0=x2-9x-5x+45

0=x(x-9)-5(x-9)

0=(x-9) (x-5)

X-9 =0. X-5= 0

X=9 or x=5

Thus, the coordinates of the points…..(9,0)or(5,0)

 

Question 9. Find out the value of a, when the distance between the points P (–3, –14) and Q (p, –5) is 9 units.

Answer 9:

Distance between the two given points (u1,v1) ( u2,v2) is :

D=√(x2-x1)²+(y2-y1)²

Distance between the points P (–3, –14) and Q (p, –5) is :

=√[(p+3)²+(-5+14)²] =9

On squaring both the L.H.S and R.H.S.

(p+3)²+81=81

(p+3)²=0

(p+3)(p+3)=0

p+3=0

p = -3

 

Question 10. Find out the point that is equidistant from the points O (–5, 4) and M (–1, 6)? How many of these points are there?

Answer 10:

Assume the point be P

As per the question,

The point P is equidistant from O (–5, 4) and M (–1, 6)

Now, the point P

= ((x1+x2)/2, (y1+y2)/2)

= ((-5-1)/2, (6+4)/2)

= (-3 , 5 )

 

Question 11. Find out the coordinates of the point Q on the x-axis that lies on the perpendicular bisector of the line segment that are joining the points A (–5, –2) and B(4, –2). Name the type of the triangle formed by the points Q, A and B.

Answer 11: Point Q is the midpoint of the side AB as the point P lies on the perpendicular bisector of the side AB.

Source: Internet

By the midpoint formula:

(x1 + x2)/2 = (-5+4)/2

= -½

X = -½

Given as the point P lies on the x axis,

Hence, y=0

P(x,y)= (-½ , 0)

Thus, it is an isosceles triangle

 

Question 12. Find out the value of m when the points (5, 1), (–2, –3) and (8, 2m) are collinear.

Answer 12:

The points A(5, 1), B(–2, –3) and C(8, 2m) are collinear.

that is Area of the ∆ABC = 0

½ [a1 (b2 – b3 ) + a2 (b3 – b1 ) + a3 (b1 – b2 )]=0

½ [5(-3 – 2m) + ( – 2)(2m – 1) + 8(1 – ( – 3))]=0

½ (-15 – 10m – 4m + 2 + 32) = 0

½ (-14m + 19) = 0

M = 19/14

 

Question 13. Find out the area for the triangle whose vertices are given as (–8, 4), (–6, 6) and (–3, 9).

Answer 13:

The given vertices are as follows:

(x₁, y₁) = (-8, 4)

(x₂, y₂) = (-6, 6)

(x₃, y₃) = (-3, 9)

Area of the triangle = (½) (x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂))

= (½) (-8(6 – 9) + -6(9 – 4) + -3(4 – 6))

= (½) (-8(-3) + -6(5) + -3(-2))

= (½) (24 – 30 + 6)

= (½) (30 – 30)

= 0 units.

 

Question 14. In what ratio does the x-axis divide the line segment that joins the points (– 4, – 6) and (–1, 7)? Find out the coordinates of the point of division.

Answer 14:

Assume the ratio in which the x-axis divides the line segment joining the points (–4, –6) and (–1, 7) = 1: k.

Now,

X-coordinate will become (-1 – 4k) / (k + 1)

Y-coordinate will become (7 – 6k) / (k + 1)

As the point P lies on the x-axis, y coordinate = 0

(7 – 6k) / (k + 1) = 0

7 – 6k = 0

K = 6/7

Then, m1 = 6 and m2 = 7

By using the section formula,

X = (m1x2 + m2x1)/(m1 + m2)

= (6(-1) + 7(-4))/(6+7)

= (-6-28)/13

= -34/13

Hence, we get,

Y = (6(7) + 7(-6))/(6+7)

= (42-42)/13

= 0

Therefore, the coordinates of point P are (-34/13, 0)

 

Question 15. When (– 4, 3) and (4, 3) are the two given vertices of an equilateral triangle, find out the coordinates of the third vertex, such that the origin lies in the interior of the triangle.

Answer 15:

Source: Internet

Assume the vertices be (x,y)

Distance between the points (x,y) & (4,3) is = √((x-4)2 + (y-3)2)……(1)

Distance between the points (x,y) & (-4,3) is = √((x+4)2 + (y-3)2)……(2)

Distance between the points (4,3) &(-4,3) is =√((4+4)2 + (3-3)2) = √(8)²=8

As per the question,

Now, the Equation (1)=(2)

(x-4)²=(x+4)²

X²-8x+16=x²+8x+16

16x=0

X=0

And the equation (1)=8

(x-4)²+(y-3)²=64……… (3)

Putting the value of x in (3)

Now, (0-4)²+(y-3)²=64

(y-3)²=64-16

(y-3)²=48

y-3=(+)4√3

y=3(+) 4√3

Neglect y = 3+4√3 as if given y = 3+4√3

Then the origin could not be the interior of the triangle

Thus, the third vertex = (0, 3-4√3)