Use app

>>Class 10
>>Maths
>> $Coordinate Geometry$

Coordinate Geometry

All

Guides

Practice

Learn

NCERT Solutions for Class 10 Maths Chapter 7 : Coordinate Geometry

Coordinate Geometry Class 10 Maths NCERT Solutions Chapter 7 is concerned with the understanding of various aspects of coordinate geometry. Students preparing for chapter 7 maths class 10 will be able to clear all their concepts on coordinate geometry at the root level. It allows us to learn geometry with algebra and understand algebra with geometry. Expert faculty of Toppr produced these solutions to assist students with their first term exam preparations. It covers all major concepts in detail, allowing students to understand the ideas better.

Class 10 Maths chapter 7 NCERT Solutions Coordinate Geometry of the section, concentrates on the essential concepts such as the distance computation between two locations in the cartesian plane, internal and external division of a line segment in a specific ratio, area calculation of a triangle, and real-world applications of these formulas. All of these solutions are designed with the new CBSE pattern in mind so that students have a complete understanding of their tests.

Class 10 Coordinate Geometry Questions and Answers are very useful for getting good grades in tests and properly preparing you with all of the important concepts. These NCERT Solutions are valuable tools that can assist you not only in covering the full syllabus but also in providing an in-depth analysis of the subjects. The NCERT Solutions for Class 10 Maths Chapter 7 are available in pdf format below, and some of them are also included in the exercises.

Table of Content

Introduction

Download PDF of NCERT Solutions for Class 10 Maths Chapter 7 : Coordinate Geometry

Access NCERT Solutions for Class 10 Maths Chapter 7 : Coordinate Geometry

Exercise 7.1

Question 1

Find the distance between the following pairs of points :
i)

(2, 3), (4, 1)

ii)

(- 5, 7), (- 1, 3)

iii)

(a, b), (- a, - b)

Medium

Solution

Verified by Toppr

(i)

= (4 - 2)^{2} + (3 - 1)^{2} = 22

units

(i i)

= (7 - 3)^{2} + (- 5 + 1)^{2} = 42

units

(i i i)

= (a + a)^{2} + (b + b)^{2} = 2 a^{2} + b^{2}

units

Question 2

Find the distance between the points

(0, 0)

and

(36, 15)

. Can you now find the distance between the two towns

A

and

B

Easy

Solution

Verified by Toppr

L e t t h e t w o p o i n t s b e P (0, 0) a n d Q (36, 15) N o w, P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} = (36 - 0)^{2} (15 - 0)^{2} = (36)^{2} + (15)^{2} = 1296 + 225 = 1521 = 39 . N o w, I t i s g i v e n t h a t B i s l o c a t e d 36 k m e a s t a n d 15 k m n o r t h o f t o w n A . L e t u s t a k e A a s o r i g i n . N o w, A B = (36 - 0)^{2} + (15 - 0)^{2} = 12596 + 225 = 1521 = 39 . H e n c e, t h e d i s tan c e b e t w e e n t h e t w o t o w n s i s 39 k m .

Question 3

Determine if the points

(1, 5), (2, 3)

and

(- 2, - 11)

are collinear , by distance formula.

Medium

Solution

Verified by Toppr

A (1, 5), B (2, 3) a n d C (- 2, - 11)

Now,

A B = (1^{2}) + (2^{2}) = 5 B C = (4^{2}) + (1 4^{2}) = 212 C A = (1^{2}) + (1 6^{2}) = 257

As,

C A > B C > A B

If points

A, B a n d C

are collinear then

A B + B C = C A

But

5 + 212 \neq = 257

So they are not collinear.

Question 4

Check whether

(5, - 2)

(6, 4)

and

(7, - 2)

are the vertices of an isosceles triangles.

Easy

Solution

Verified by Toppr

Let ABC be the triangle

A (5, - 2), B (6, 4), C (7, - 2)

By distance formula

A B = (5 - 6)^{2} + (- 2 - 4)^{2} = (- 1)^{2} + (- 6)^{2} 37

B C = (6 - 7)^{2} + (4 - (- 2)^{2} = (- 1)^{2} + (6)^{2} = 37

C A = (7 - 5)^{2} + (- 2 - (- 2))^{2} = (2)^{2} + 0 = 2

Since

A B = B C = 37

Hence

(5, - 2), (6, 4)

(7, - 2)

are vertices of isosceles triangle.

Question 5

In a classroom,

4

friends are seated at the points ,

A, B, C

and

D

as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, "Don't you think

A B C D

is a square " ? Chameli disagrees. Using distance formula, find which of them is correct.

Medium

Solution

Verified by Toppr

From the figure co-ordinates of points can be expressed as shown below,

A (3, 4), B (6, 7), C (9, 4), D (6, 1)

A B = (6 - 3)^{2} + (7 - 4)^{2} = 3^{2} + 3^{2}

= 9 + 9 = 18 = 32

B C = (9 - 6)^{2} + (6 - 7)^{2} = 3^{2} + (- 3)^{2}

= 9 + 9 = 18 = 32

C D = (6 - 9)^{2} + (1 - 4)^{2} = (- 3)^{2} + (- 3)^{2}

= 9 + 9 = 18 = 32

D A = (6 - 3)^{2} + (1 - 4)^{2} = 3^{2} + (- 3)^{2}

= 9 + 9 = 18 = 32

Four sides of the quadrilateral

A B C D

are equal

A B = B C = C D = D A = 32

D i a g o n a l A C = (9 - 3)^{2} + (4 - 4)^{2} = 6

D i a g o n a l B D = (6 - 6)^{2} + (1 - 7)^{2} = 6

D i a g o n a l A C = D i a g o n a l B D

Therefore

A B C D

is a square

Therefore, Champa is correct between two

Question 6

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i)

(- 1, - 2), (1, 0), (- 1, 2), (- 3, 0)

(ii)

(- 3, 5), (3, 1), (0, 3), (- 1, - 4)

(iii)

(4, 5), (7, 6), (4, 3), (1, 2)

Medium

Solution

Verified by Toppr

(i) Let the given points are

A (- 1, - 2)

B (1, 0)

C (- 1, 2)

and

D (- 3, 0)

Then,

A B = (1 + 1)^{2} + (0 + 2)^{2} = 2^{2} + 2^{2} = 4 + 4 = 8

B C = (- 1 - 1)^{2} + (2 - 0)^{2} = (2^{2} + 2^{2}) = 4 + 4 = 8

C D = ((- 3) - (- 1))^{2} + (0 - 2)^{2} = 2^{2} + (- 2)^{2} = 4 + 4 = 8

D A = (- 3) - (- 1))^{2} + (0 - (- 2))^{2} = (- 2)^{2} + 2^{2} = 4 + 4 = 8

A C ((- 1) - (- 1))^{2} + (2 - (- 2))^{2} = 0 + 4^{2} = 16 = 4

B D = (- 3 - 1)^{2} + (0 - 0)^{2} = - 4^{2} = 16 = 4

Since the four sides

A B, B C, C D

and

D A

are equal and the diagonals

A C

and

B D

are equal .

∴

Quadrilateral

A B C D

is a square.

(ii)Let the given points are

A (- 3, 5)

B (3, 1)

C (0, 3)

and

D (- 1, - 4)

Then

A B = (- 3 - 3)^{2} + (5 - 1)^{2} = (- 6)^{2} + 4^{2} = 36 + 16 = 52

B C = (3 - 0)^{2} + (1 - 3)^{2} = (3^{2} + (- 2) 2^{2}) = 9 + 4 = 11

C D = (0 - (- 1))^{2} + (3 - (- 4))^{2} = 1^{2} + (7)^{2} = 1 + 49 = 50

D A = (- 1) - (- 3))^{2} + ((- 4) - 5))^{2} = (2)^{2} + (- 9)^{2} = 4 + 81 = 85

Here

A B \neq = B C \neq = C D \neq = D A

∴

it is a quadrilateral.

(iii)Let the given points are

A (4, 5)

B (7, 6)

C (4, 3)

and

D (1, 2)

Then

A B = (7 - 4)^{2} + (6 - 5)^{2} = 3^{2} + 1^{2} = 9 + 1 = 10

B C = (4 - 7)^{2} + (3 - 6)^{2} = ((- 3)^{2} + (- 3)^{2}) = 9 + 9 = 18

C D = (1 - 4)^{2} + (2 - 3)^{2} = (- 3)^{2} + (- 1)^{2} = 9 + 1 = 10

D A = (1 - 4)^{2} + (2 - 5)^{2} = (- 3)^{2} + (- 3)^{2} = 9 + 9 = 18

A C (4 - 4)^{2} + (3 - 5)^{2} = 0 + (- 2)^{2} = 4 = 2

B D = (1 - 7)^{2} + (2 - 6)^{2} = (- 6)^{2} + (- 4)^{2} = 36 + 16 = 52

Here

A B = C D, B C = D A

. But

A C \neq = B D

Hence the pairs of opposite sides are equal but diagonal are not equal so it is a parallelogram.

Question 7

Point on the

x

-axis which is equidistant from

(2, - 5)

and

(- 2, 9)

is:

(- 7, 0)

(- 14.0)

(7, 0)

(14, 0)

Medium

Solution

Verified by Toppr

Since point on

x -

axis, then coordinate of the point is

(x, 0)

.
According to the question this point

(x, 0)

is equidistant from the points

(2, - 5)

and

(- 2, 9)

.
That is, distance from

(x, 0)

and

(2, - 5)

=

distance from

(x, 0)

and

(- 2, 9)

(2 - x)^{2} + (- 5 - 0)^{2} = (- 2 - x)^{2} + (9 - 0)^{2} \Rightarrow (2 - x)^{2} + (- 5)^{2} = (- 2 - x)^{2} + (9)^{2} \Rightarrow 4 - 4 x + x^{2} + 25 = 4 + 4 x + 4 + 81 \Rightarrow 8 x = 25 - 81 \Rightarrow 8 x = - 56 \Rightarrow x = - 7

So, point is

(- 7, 0)

Question 8

Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.

8, 2

-9, 3

-9, 5

-8 , 2

Medium

Solution

Verified by Toppr

P Q = 10

\Rightarrow (10 - 2)^{2} + (y + 3)^{2} = 10

64 + y^{2} + 9 + 6 y = 10

Squaring both sides,

y^{2} + 6 y - 27 = 0

\Rightarrow y^{2} + 9 y - 3 y - 37 = 0

\Rightarrow y (y + 9) - 3 (y + 9) = 0

\Rightarrow (y + 9) (y - 3) = 0

∴ y = - 9, 3

Question 9

Q (0, 1)

is equidistant from

P (5, - 3)

and

R (x, 6)

, find the values of

x

. Also find the distances

Q R

and

P R

Medium

Solution

Verified by Toppr

As given

Q (0, 1)

is equidistant from

P (5, - 3)

and

R (x, 6)

\Rightarrow P Q = Q R

\Rightarrow (0 - 5)^{2} + (1 - (- 3))^{2}

= (x - 0)^{2} + (1 - 6)^{2}

[ By using disatnce formula

= (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

]

\Rightarrow 25 + 16 = x^{2} + 25

\Rightarrow 41 = x^{2} + 25

\Rightarrow x^{2} = 16

\Rightarrow x = 4

Distance between

Q (0, 1)

and

R (4, 6)

Q R = (4 - 0)^{2} + (6 - 1)^{2} = 4^{2} + 5^{2} = 16 + 25 = 41

Distance between

P (5, - 3)

and

R (4, 6)

P R = (4 - 5)^{2} + (6 + 3)^{2} = 1 + 81 = 82

Question 10

Find a relation between

x

and

y

such that the point

(x, y)

is equidistant from the point

(3, 6)

and

(- 3, 4)

Medium

Solution

Verified by Toppr

Let the point

P (x, y)

is equidistant from the points

A (3, 6)

and

B (- 3, 4)

∴ P A = P B

By using the distance formula

= (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

we have

\Rightarrow (x - 3)^{2} + (y - 6)^{2} = (x + 3)^{2} + (y - 4)^{2}

\Rightarrow (x - 3)^{2} + (y - 6)^{2} = (x + 3)^{2} + (y - 4)^{2}

\Rightarrow x^{2} - 6 x + 9 + y^{2} - 12 y + 36 = x^{2} + 6 x + 9 + y^{2} - 8 y + 16

\Rightarrow - 6 x - 6 x - 12 y + 8 y + 36 - 16 = 0

\Rightarrow - 12 x - 4 y + 20 = 0

\Rightarrow 3 x + y - 5 = 0

Hence this is a relation between

x

and

y

Practice more questions

Fast Revision

NCERT Solutions for Class 10 Maths Chapter 7 : Coordinate Geometry

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry – Brief Overview

1. Coordinates

The x-coordinate, or abscissa, of a point, is its distance from the y-axis. The y-coordinate, or ordinate, of a point, is its distance from the x-axis. A point on the x-axis has coordinates of the form (x, 0), while a point on the y-axis has coordinates of the form (0, y).

2. Coordinate Geometry

Coordinate geometry is the branch of mathematics that uses algebraic methods to solve geometrical issues. It is the study of cartesian coordinates. This system is typically used to alter the equations of two-dimensional shapes such as triangles, squares, circles, and so on.

Coordinate geometry is significant in mathematics because it allows us to identify points on any given plane. Coordinate geometry is also known as the study of graphs. It also has numerous applications in trigonometry, dimensional geometry, calculus, and other sciences.

3. Distance Formula

The distance between any two points, P(x₁, y₁) and Q(x₂, y₂) in the plane are given by:

Also, the distance between P(x₁, y₁) and the origin is:

4. Section Formula

The section formula assists us in determining the coordinates of the point that divides the line in the ratio m:n. If P is the point splitting the line AB internally in the ratio m:n where A(x₁, y₁) and B(x₂, y₂).

P's coordinates will be:

If P is the point splitting the line AB externally in the ratio m:n where A(x₁, y₁) and B(x₂, y₂).

P's coordinates will be:

5. Area of Triangle

The area of ∆ABC formed by the vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) is given by:

Area of Triangle = ½ x [x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ – y₂)]

Related Chapters

Chapter 1 : Real Numbers

Chapter 2 : Polynomials

Chapter 3 : Pair of Linear Equations in Two Variables

Chapter 4 : Quadratic Equations

Chapter 5 : Arithmetic Progression

Chapter 6 : Triangles

Chapter 8 : Introduction to Trigonometry

Chapter 9 : Some Applications of Trigonometry

Chapter 10 : Circles

Chapter 11 : Constructions

Chapter 12 : Area Related to Circles

Chapter 13 : Surface Areas and Volumes

Chapter 14 : Statistics

Chapter 15 : Probability

Frequently Asked Questions on NCERT Class 10 Maths Chapter 7 : Coordinate Geometry

Q1. What are the most important topics in NCERT Class 10 Chapter 7 Coordinate Geometry?

Answer: The distance formula, the section formula, and the area of the triangle are all important concepts in class 10 chapter 7 Coordinate Geometry. Questions on all three topics carry a considerable weightage in tests, therefore students should practice them thoroughly.

Q2. How many exercises are there in Chapter 7 of Class 10 Maths?

Answer: There are four exercises in total. Class 10 Maths Chapter 7 Coordinate Geometry has 33 problems, 20 of which are pretty easy, 7 of which are moderately tough, and 6 of which are long answer intricate sums. These questions are spread among four exercises in this chapter. They are both theoretical and practical in character, based on the cartesian plane and the associated formulas.

Q3. What is the purpose of Coordinate Geometry?

Answer: Coordinate geometry has many uses in everyday life. It is an important part of mathematics that helps us locate points on a plane. Furthermore, it has several applications in trigonometry, calculus, dimensional geometry, and other sciences. A few examples of coordinate geometry applications are: GPS positions in digital maps to pinpoint exact locations, Coordinate geometry is also employed in the construction of highways, buildings, and other structures, and Coordinate geometry is also employed in the cartographic representation of 2D and 3D objects.

Related Chapters

Chapter 1 : Real Numbers

Chapter 2 : Polynomials

Chapter 3 : Pair of Linear Equations in Two Variables

Chapter 4 : Quadratic Equations

Chapter 5 : Arithmetic Progression

Chapter 6 : Triangles

Chapter 8 : Introduction to Trigonometry

Chapter 9 : Some Applications of Trigonometry

Chapter 10 : Circles

Chapter 11 : Constructions

Chapter 12 : Area Related to Circles