Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.1 - Coordinate Geometry

Last updated date: 19th Jul 2024
Total views: 600.3k
Views today: 8.00k

## NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1 - FREE PDF Download

The NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1 on Coordinate Geometry offers detailed answers to the exercises given. These solutions are designed to help students prepare for their CBSE Class 10 board exams. It's important for students to go through these solutions carefully as they cover various types of questions related to Coordinate Geometry. By practicing with these solutions, students can enhance their understanding and be better equipped to tackle similar questions in their Class 10 board exams.

Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 7 Exercise 7.1 Class 10 | Vedantu
3. Access PDF for Maths NCERT Chapter 7 Coordinate Geometry Exercise 7.1 Class 10
4. Class 10 Maths Chapter 7: Exercises Breakdown
5. Other Study Materials of CBSE Class 10 Maths Chapter 7
6. A Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 7 Exercise 7.1 Class 10 | Vedantu

• In this article, students can enhance their problem-solving skills and develop a deeper insight into coordinate geometry concepts.

• Understanding the Cartesian plane with X and Y axes, and representing points using coordinates (x, y).

• Applying the distance formula to find the distance between two points on the coordinate plane.

• Plotting points on the graph based on their coordinates.

• Finding the distance between points with given coordinates.

• Determining the coordinates of a point which is equidistant from two other points.

• Exercise 7.1 class 10 NCERT solutions has over all 10 questions.

Competitive Exams after 12th Science
Watch videos on
NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.1 - Coordinate Geometry
COORDINATE GEOMETRY in One Shot (Complete Chapter) | CBSE 10 Maths Chapter 6 - Term 1 Exam | Vedantu
Vedantu 9&10
Subscribe
Share
6.1K likes
117.2K Views
2 years ago
COORDINATE GEOMETRY L-1 [Distance Formula] CBSE Class 10 Maths Chapter 7 (Term 1 Exam) NCERT Vedantu
Vedantu 9&10
8.1K likes
115K Views
2 years ago
Coordinate Geometry in One Shot | CBSE Class 10 Maths Chapter 7 NCERT Solutions | Vedantu
Vedantu 9&10
11.1K likes
314.5K Views
3 years ago
Coordinate Geometry L1 | Distance Formula | CBSE Class 10 Maths | NCERT Solutions | Umang Vedantu
Vedantu 9&10
9K likes
216.9K Views
4 years ago

## Access PDF for Maths NCERT Chapter 7 Coordinate Geometry Exercise 7.1 Class 10

1. Find the distance between the following pairs of points:

(i) $\left( \text{2,3} \right)\text{,}\left( \text{4,1} \right)$

Ans: The given points are $\text{A}\left( \text{2, 3} \right)$ and $\text{B}\left( \text{4, 1} \right)$

Using the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$

By comparing the points with the formula we get, ${{\text{x}}_{\text{1}}}\text{=2}$, ${{\text{y}}_{\text{1}}}\text{=3}$, ${{\text{x}}_{\text{2}}}\text{=4}$, ${{\text{y}}_{\text{2}}}\text{=1}$

Distance between $\text{A}\left( \text{2,3} \right)$ and $\text{B}\left( \text{4,1} \right)$ is given by,

AB = $\sqrt{{{\text{(2-4)}}^{\text{2}}}\text{+(3-1}{{\text{)}}^{\text{2}}}}$

= $\sqrt{{{\text{(-2)}}^{\text{2}}}\text{+(2}{{\text{)}}^{\text{2}}}}$

= $\sqrt{\text{4+4}}$

= $\sqrt{\text{8}}$

= $\text{2}\sqrt{\text{2}}$

(ii) $\left( -\mathbf{5},\mathbf{7} \right),\text{ }\left( -\mathbf{1},\mathbf{3} \right)$

Ans: The given points are $\text{C}\left( \text{-5, 7} \right)$ and $\text{D}\left( \text{-1, 3} \right)$

Using the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$

By comparing the points with the formula we get, ${{\text{x}}_{\text{1}}}\text{= -5}$, ${{\text{y}}_{\text{1}}}\text{= 7}$, ${{\text{x}}_{\text{2}}}\text{= -1}$, ${{\text{y}}_{\text{2}}}\text{= 3}$

Distance between $C\left( \text{-5,7} \right)$and$D\left( \text{-1,3} \right)$ is given by,

CD  = $\sqrt{{{\text{(-5-(-1))}}^{\text{2}}}\text{+(7-3}{{\text{)}}^{\text{2}}}}$

= $\sqrt{{{\text{(-4)}}^{\text{2}}}\text{+(4}{{\text{)}}^{\text{2}}}}$

= $\sqrt{\text{16+16}}$

= $\sqrt{\text{32}}$

= $\text{4}\sqrt{\text{2}}$

(iii) $\left( \mathbf{a},\mathbf{b} \right),\left( -\mathbf{a},-\mathbf{b} \right)$

Ans: The given points are $\text{X}\left( \text{a, b} \right)$ and $\text{Y}\left( \text{-a, -b} \right)$

Using the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$

By comparing the points with the formula we get, ${{\text{x}}_{\text{1}}}\text{= a}$, ${{\text{y}}_{\text{1}}}\text{= b}$, ${{\text{x}}_{\text{2}}}\text{= -a}$,${{\text{y}}_{\text{2}}}\text{= -b}$

Distance between $\text{X}\left( \text{a, b} \right)$ and $\text{Y}\left( \text{-a, -b} \right)$ is given by,

XY = $\sqrt{{{\text{(a-(-a))}}^{\text{2}}}\text{+(b-(-b)}{{\text{)}}^{\text{2}}}}$

= $\sqrt{{{\text{(2a)}}^{\text{2}}}\text{+(2b}{{\text{)}}^{\text{2}}}}$

= $\sqrt{\text{4}{{\text{a}}^{\text{2}}}\text{+4}{{\text{b}}^{\text{2}}}}$

= $\text{2}\sqrt{{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}}$

2. Find the distance between the points $\left( \text{0,0} \right)$ and $\left( \text{36,15} \right)$. Can you now find the distance between the two towns A and B discussed in Section 7.2?

Ans: According to the , the given points are $\text{P}\left( \text{0,0} \right)$ and $\text{Q}\left( \text{36,15} \right)$.

Using the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$

By comparing the points with the formula we get, ${{\text{x}}_{\text{1}}}\text{=0}$, ${{\text{y}}_{\text{1}}}\text{=0}$, ${{\text{x}}_{\text{2}}}\text{=36}$, ${{\text{y}}_{\text{2}}}\text{=15}$.

The distance between the points $\text{P}\left( \text{0,0} \right)$ and $\text{Q}\left( \text{36,15} \right)$ is given by,

PQ  = $\sqrt{{{\text{(36-0)}}^{\text{2}}}\text{+(15-0}{{\text{)}}^{\text{2}}}}$

= $\sqrt{\text{3}{{\text{6}}^{\text{2}}}\text{+1}{{\text{5}}^{\text{2}}}}$

= $\sqrt{\text{1296+225}}$

= $\sqrt{\text{1521}}$

= $\text{39}$

Yes, we can find the distance between two towns A and B using the distance formula only. Two town are situated at point  (0,0) and (36,15) respectively. So, the distance between both town will be 39 km.

3. Determine if the points $\left( \text{1,5} \right)\text{, }\left( \text{2,3} \right)$ and $\left( \text{-2,-11} \right)$ are collinear.

Ans: Three points are collinear if and only if they are on the same line.

Let the given points are $A\left( \text{1,5} \right)\text{, B}\left( \text{2,3} \right)$ and $C\left( \text{-2,-11} \right)$

Using the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$,

$AB=\sqrt{{{(1-2)}^{2}}+{{(5-3)}^{2}}}$

= $\sqrt{{{\text{1}}^{\text{2}}}\text{+}{{\text{2}}^{\text{2}}}}$

= $\sqrt{\text{5}}$

$B\text{C}=\sqrt{{{(2-(-2))}^{2}}+{{(3-(-11))}^{2}}}$

= $\sqrt{\text{16+196}}$

= $\sqrt{\text{212}}$

$\text{AC}=\sqrt{{{(1-(-2))}^{2}}+{{(5-(-11))}^{2}}}$

= $\sqrt{\text{9+256}}$

= $\sqrt{265}$

Here,$\text{AB+AC}\ne \text{BC}$ and $\text{BC+AC}\ne AB$ and $\text{AB+BC}\ne A\text{C}$. Since, the sum of distance of any two point is not equal to third point.

Therefore, the points $\left( \text{1,5} \right)\text{, }\left( \text{2,3} \right)$ and $\left( \text{-2,-11} \right)$ are not collinear.

4. Check whether $\left( \text{5,-2} \right)\text{, }\left( \text{6,4} \right)$ and $\left( \text{7,-2} \right)$ are the vertices of an isosceles triangle.

Ans: To prove the points to be an isosceles triangle the points must be collinear and any two sides must be of equal length.

Let the points $\left( \text{5,-2} \right)\text{, }\left( \text{6,4} \right)$ and $\left( \text{7,-2} \right)$ are representing the vertices

A, B, and C of the given triangle respectively.

Using the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$,

$AB=\sqrt{{{(5-6)}^{2}}+{{(-2-4)}^{2}}}$

= $\sqrt{{{\text{(-1)}}^{2}}\text{+(-6}{{\text{)}}^{2}}}$

= $\sqrt{\text{1+36}}$

= $\sqrt{37}$

$BC=\sqrt{{{(6-7)}^{2}}+{{(4-(-2))}^{2}}}$

= $\sqrt{{{\text{(-1)}}^{2}}\text{+(6}{{\text{)}}^{2}}}$

= $\sqrt{\text{1+36}}$

= $\sqrt{37}$

$AC=\sqrt{{{(5-7)}^{2}}+{{(-2-(-2))}^{2}}}$

= $\sqrt{{{\text{(-2)}}^{2}}\text{+(0}{{\text{)}}^{2}}}$

= $\sqrt{4}$

= 2

Here, AB = BC. As two sides are equal in length and the sum of distance of any two point is not equal to third point. Therefore, ABC is an isosceles triangle.

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

Ans: From the above figure, the coordinates of A, B, C, D are$\left( \text{3,4} \right)\text{, }\left( \text{6,7} \right)\text{, }\left( \text{9,4} \right)$ and $\left( \text{6,1} \right)$ respectively.

Using the Distance formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$

$AB=\sqrt{{{(3-6)}^{2}}+{{(4-7)}^{2}}}$

= $\sqrt{{{\text{(-3)}}^{2}}\text{+(-3}{{\text{)}}^{2}}}$

= $\sqrt{\text{9+9}}$

= $\sqrt{18}$

= $3\sqrt{2}$

$BC=\sqrt{{{(6-9)}^{2}}+{{(7-4)}^{2}}}$

= $\sqrt{{{\text{(-3)}}^{2}}\text{+}{{\text{3}}^{2}}}$

= $\sqrt{9+9}$

= $\sqrt{18}$

=$3\sqrt{2}$

$CD=\sqrt{{{(9-6)}^{2}}+{{(4-1)}^{2}}}$

= $\sqrt{{{\text{3}}^{2}}\text{+}{{\text{3}}^{2}}}$

= $\sqrt{9+9}$

= $\sqrt{18}$

= $3\sqrt{2}$

$AD=\sqrt{{{(3-6)}^{2}}+{{(4-1)}^{2}}}$

= $\sqrt{{{\text{(-3)}}^{2}}\text{+}{{\text{3}}^{2}}}$

= $\sqrt{9+9}$

= $\sqrt{18}$

= $3\sqrt{2}$

Diagonal $AC=\sqrt{{{(3-9)}^{2}}+{{(4-4)}^{2}}}$

= $\sqrt{{{\text{(-6)}}^{2}}\text{+}{{\text{0}}^{2}}}$

= $\text{6}$

Diagonal $BD=\sqrt{{{(6-6)}^{2}}+{{(7-1)}^{2}}}$

= $\sqrt{{{0}^{2}}\text{+}{{\text{6}}^{2}}}$

= $\text{6}$

Here, all four sides AB, BC, CD, and AD are of same length and diagonals AC and BD are also of equal length . Therefore, we can conclude that ABCD is a square and hence, Champa was correct.

6. Name the type of quadrilateral formed, if any, by the following points, and

i. $\text{(-1,-2), (1,0), (-1,2), (-3,0)}$

Ans: Let the points $\text{(-1,-2), (1,0), (-1,2), (-3,0)}$be representing the vertices A, B, C, and D of the given quadrilateral respectively.

Using the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$,

$AB=\sqrt{{{(-1-1)}^{2}}+{{(-2-0)}^{2}}}$

$=\sqrt{{{(-2)}^{2}}\text{+(-2}{{\text{)}}^{2}}}$

$=\sqrt{\text{4+4}}$

$=\sqrt{8}\text{ }$

= $\text{2}\sqrt{2}$

$BC=\sqrt{{{(1-(-1))}^{2}}+{{(0-2)}^{2}}}$

$=\sqrt{{{(2)}^{2}}\text{+(-2}{{\text{)}}^{2}}}$

$=\sqrt{\text{4+4}}$

$=\sqrt{8}$

= $\text{2}\sqrt{2}$

$CD=\sqrt{{{(-1-(-3))}^{2}}+{{(2-0)}^{2}}}$

$=\sqrt{{{2}^{2}}\text{+}{{\text{2}}^{2}}}$

$=\sqrt{\text{4+4}}$

$=\sqrt{8}$

= $\text{2}\sqrt{2}$

$AD=\sqrt{{{(-1-(-3))}^{2}}+{{(-2-0)}^{2}}}$

$=\sqrt{{{2}^{2}}\text{+}{{\text{2}}^{2}}}$

$=\sqrt{\text{4+4}}$

$=\sqrt{8}$

= $\text{2}\sqrt{2}$

Diagonal $AC=\sqrt{{{(-1-(-1))}^{2}}+{{((-2)-2)}^{2}}}$

$=\sqrt{{{0}^{{}}}\text{+}{{\text{4}}^{2}}}$

$=\sqrt{16}$

$\text{= 4}$

Diagonal $BD=\sqrt{{{(1-(-3))}^{2}}+{{(0-0)}^{2}}}$

$=\sqrt{{{\text{4}}^{2}}+0}$

$=\sqrt{16}$

$\text{= 4}$

Therefore, AB = BC = CD = AD

Here, all the sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.

ii. $\text{(-3,5), (3,1), (0,3), (-1,-4)}$

Ans: Let the points $\text{(-3,5), (3,1), (0,3), (-1,-4)}$ be representing the vertices A, B, C, and D of the given quadrilateral respectively.

Using the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$,

$AB=\sqrt{{{(-3-3)}^{2}}+{{(5-1)}^{2}}}$

$=\sqrt{{{(-6)}^{2}}\text{+(4}{{\text{)}}^{2}}}$

$=\sqrt{36+16}$

$=\sqrt{52}$

$\text{=2}\sqrt{13}$

$BC=\sqrt{{{(3-0)}^{2}}+{{(1-3)}^{2}}}$

$=\sqrt{{{3}^{2}}\text{+(-2}{{\text{)}}^{2}}}$

$=\sqrt{\text{9+4}}$

$=\sqrt{13}$

$CD=\sqrt{{{(0-(-1))}^{2}}+(3-{{(-4)}^{2}}}$

$=\sqrt{{{1}^{2}}\text{+}{{\text{7}}^{2}}}$

$=\sqrt{\text{1+49}}$

$=\sqrt{50}$

$\text{=5}\sqrt{2}$

$AD=\sqrt{{{(-3-(-1))}^{2}}+(5-{{(-4)}^{2}}}$

$=\sqrt{{{(-2)}^{2}}\text{+}{{\text{9}}^{2}}}$

$=\sqrt{\text{4+81}}$

$=\sqrt{85}$

Diagonal $AC=\sqrt{{{(0-(-3))}^{2}}+{{(3-5)}^{2}}}$

$=\sqrt{{{3}^{2}}\text{+(-2}{{\text{)}}^{2}}}$

$=\sqrt{\text{9+4}}$

$=\sqrt{13}$

Diagonal $BD=\sqrt{{{(-1-3)}^{2}}+{{(-4-1)}^{2}}}$

$=\sqrt{{{(-4)}^{2}}+{{(-5)}^{2}}}$

$=\sqrt{16+25}$

$\text{= }\sqrt{41}$

Here, all the sides of this quadrilateral are of different length i.e.,$\text{AB}\ne \text{BC}\ne \text{AC}\ne \text{AD}$ and also AC + BC = AB which means that A, B, C are collinear. Therefore, no quadrilateral can be formed from these points.

iii. $\text{(4,5), (7,6), (4,3), (1,2)}$

Ans: Let the points $\text{(4,5), (7,6), (4,3), (1,2)}$ be representing the vertices A, B, C, and D of the given quadrilateral respectively.

Using the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$,

$AB=\sqrt{{{(4-7)}^{2}}+{{(5-6)}^{2}}}$

$=\sqrt{{{(-3)}^{2}}\text{+(-1}{{\text{)}}^{2}}}$

$=\sqrt{9+1}$

$=\sqrt{10}$

$BC=\sqrt{{{(7-4)}^{2}}+{{(6-3)}^{2}}}$

$=\sqrt{{{3}^{2}}\text{+}{{\text{3}}^{2}}}$

$=\sqrt{\text{9+9}}$

$=\sqrt{18}$

$CD=\sqrt{{{(4-1)}^{2}}+{{(3-2)}^{2}}}$

$=\sqrt{{{3}^{2}}\text{+}{{\text{1}}^{2}}}$

$=\sqrt{\text{9+1}}$

$=\sqrt{10}$

$AD=\sqrt{{{(4-1)}^{2}}+{{(5-2)}^{2}}}$

$=\sqrt{{{3}^{2}}\text{+}{{\text{3}}^{2}}}$

$=\sqrt{\text{9+9}}$

$=\sqrt{18}$

Diagonal $AC=\sqrt{{{(4-4)}^{2}}+{{(5-3)}^{2}}}$

$=\sqrt{0\text{+}{{\text{2}}^{2}}}$

$=\sqrt{4}\text{ = 2}$

Diagonal $BD=\sqrt{{{(7-1)}^{2}}+{{(6-2)}^{2}}}$

$=\sqrt{{{6}^{2}}+{{4}^{2}}}$

$=\sqrt{36+16}\text{ = }\sqrt{52}$

$=2\sqrt{13}$

Here, the opposite sides of this quadrilateral are of the same length i.e., AB = CD and BC = AD. But, the diagonals are of different lengths.

Therefore, the given points are the vertices of a parallelogram.

7. Find the point on the x-axis which is equidistant from $\text{(2,-5)}$ and $\text{(-2,9)}$.

Ans: If the point is on x-axis the co-ordinates will be $P\left( \text{x, 0} \right)$. We have to find a point on x-axis which is equidistant from $A\left( \text{2,-5} \right)$ and $B\left( \text{-2,9} \right)$.

Using the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$

$PA=\sqrt{{{(x-2)}^{2}}+{{(0-(-5))}^{2}}}$

$=\sqrt{{{(x-2)}^{2}}+{{5}^{2}}}$

$PB=\sqrt{{{(x-(-2))}^{2}}+{{(0-(-9))}^{2}}}$

$=\sqrt{{{(x+2)}^{2}}+{{9}^{2}}}$

By the given condition, these distances are equal in measure.

Hence PA = PB

$\sqrt{{{(x-2)}^{2}}+{{5}^{2}}}=\sqrt{{{(x+2)}^{2}}+{{9}^{2}}}$

$\Rightarrow {{(x-2)}^{2}}+25={{(x+2)}^{2}}+81$

$\Rightarrow {{\text{x}}^{2}}+4-4x+25={{x}^{2}}+4+4x+81$

$\Rightarrow \text{8x= 25-81}$

$\Rightarrow \text{8x= -56}$

$\Rightarrow \text{x= -7}$

Hence, $P\left( \text{-7, 0} \right)$ is the point on the x-axis which is equidistance from the given points.

8. Find the values of y for which the distance between the points $\text{P(2,-3)}$ and $\text{Q(10,y)}$ is $\text{10}$ units.

Ans: We know that the distance between the two points is given by the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$

Distance between points $A\left( \text{2,-3} \right)$ and $B\left( \text{10,y} \right)$ is $\text{10}$ units. Therefore, by substituting the values of points A (2, −3) and B (10, y) in Distance Equation

$\sqrt{{{\text{(2-10)}}^{\text{2}}}\text{+(-3-y}{{\text{)}}^{\text{2}}}}=10$

$\Rightarrow \sqrt{{{\text{(-8)}}^{\text{2}}}\text{+(3+y}{{\text{)}}^{\text{2}}}}=10$

Squaring both sides

$\Rightarrow \text{64+(3+y}{{\text{)}}^{\text{2}}}=100$

$\Rightarrow {{\text{(y+3)}}^{\text{2}}}=36$

$\Rightarrow \text{(y+3)}=\pm 6$

$\Rightarrow \text{(y+3)}=6$ or $\text{(y+3)}=-6$

Therefore, $\text{y = -3 or 9}$ are the possible values for y.

9. If $\text{Q(0,1)}$ is equidistant from $\text{P(5,-3)}$ and $\text{R(x,6)}$, find the values of x. Also find the distance QR and PR.

Ans: We know that the distance between the two points is given by the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$

As $\text{Q(0,1)}$ is equidistant from $\text{P(5,-3)}$ and $\text{R(x,6)}$.

So, PQ = QR

$\sqrt{{{(5-0)}^{2}}+{{(-3-1)}^{2}}}=\sqrt{{{(0-x)}^{2}}+{{(1-6)}^{2}}}$

$\Rightarrow \sqrt{{{5}^{2}}+{{(-4)}^{2}}}=\sqrt{{{(-x)}^{2}}+{{(-5)}^{2}}}$

$\Rightarrow 25+16={{x}^{2}}+25$      (By squaring both sides)

$\Rightarrow 16={{x}^{2}}$

$\Rightarrow {{x}^{2}}=\pm 4$

Therefore, point R is $\text{(4,6)}$ or $\text{(-4,6)}$

Case (1),

When point R is $\text{(4,6)}$,

Distance between $\text{P(5,-3)}$ and $\text{R(4,6)}$

PR=$\sqrt{{{(5-4)}^{2}}+{{(-3-6)}^{2}}}$

$=\sqrt{{{(1)}^{2}}+{{(-9)}^{2}}}$

$=\sqrt{1+81}$

$=\sqrt{82}$

Distance between $\text{Q(0,1)}$ and $\text{R(4,6)}$

QR = $\sqrt{{{(0-4)}^{2}}+{{(1-6)}^{2}}}$

$=\sqrt{{{(-4)}^{2}}+{{(-5)}^{2}}}$

$=\sqrt{16+25}$

$=\sqrt{41}$

Case (2),

When point R is $\text{(-4,6)}$,

Distance between $\text{P(5,-3)}$ and $\text{R(-4,6)}$

PR = $\sqrt{{{(5-(-4))}^{2}}+{{(-3-6)}^{2}}}$

$=\sqrt{{{(9)}^{2}}+{{(-9)}^{2}}}$

$=\sqrt{81+81}$

$=9\sqrt{2}$

Distance between $\text{Q(0,1)}$ and $\text{R(-4,6)}$

QR =$\sqrt{{{(0-(-4))}^{2}}+{{(1-6)}^{2}}}$

$=\sqrt{{{(4)}^{2}}+{{(-5)}^{2}}}$

$=\sqrt{16+25}$

$=\sqrt{41}$

10. Find a relation between x and y such that the point (x, y) is equidistant from the point $\text{(3,6)}$ and $\text{(-3,4)}$.

Ans: We know that the distance between the two points is given by the Distance Formula = $\sqrt{{{\text{(}{{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}}\text{)}}^{\text{2}}}\text{+(}{{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$

Point P(x, y) be equidistant from points $\text{A(3,6)}$ and $\text{B(-3,4)}$.

Therefore, PA = PB

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

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

Squaring both sides

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

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

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

$\Rightarrow 20=12x+4y$

$\Rightarrow 3x+y=5$

$\Rightarrow 3x+y-5=0$

Thus, the relation between x and y is given by $3x+y-5=0$

## Conclusion

Class 10 Maths Chapter 7 Exercise 7.1 - Coordinate Geometry, is crucial for a solid foundation in math. This exercise delves into the realm of coordinate geometry, in mathematics. Students can grasp the fundamental principles of coordinate geometry and its applications. Regular practice with NCERT solutions provided by platforms like Vedantu can enhance comprehension and problem-solving skills. Pay attention to the step-by-step solutions provided, grasp the underlying principles, and ensure clarity on the concepts before moving forward. Consistent practice and understanding will lead to proficiency in Coordinate Geometry.

## Class 10 Maths Chapter 7: Exercises Breakdown

 Exercise Number of Questions Exercise 7.2 10 Questions & Solutions (2 Short Answers, 8 Long Answer)

## A Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.1 - Coordinate Geometry

1. What will I learn in this chapter?

Coordinate Geometry of NCERT Class 10 Maths syllabus will teach you how to find the distance between points with the given coordinates and you shall learn about points on a cartesian plane, distance formula, section formula, midpoint, points of trisection, the centroid of a triangle and areas from coordinates. NCERT Solutions for Class 10 Maths Chapter 7 consist of 4 exercises which have 33 questions and they cover all the topics from the chapter. These solutions are also drafted in a step by step way to make it easy for the students. In NCERT Class 10 Maths Chapter 7, you will be taught about the distance between the two points from the given coordinates. You shall see how the area of the triangle is formed with the three given points.

2. Explain about the area of a triangle, polygon and the centroid of a triangle.

Area of a Triangle:

In this part of Coordinate Geometry from Class 10 Maths Chapter 7 of NCERT Solutions, you will be solving the questions based on the below-mentioned formula.

Area of a trapezium = ½ (Sum of parallel sides) (distance between them)

The Centroid of a Triangle:

In this section of Coordinate Geometry of Class 10 Maths, you will be learning about what is the centroid of a triangle and why the three medians of the triangle meet are known as the centroid of a triangle.

Area of a Polygon:

In this segment of NCERT CBSE Class 10 Maths Chapter 7, you will be learning how to find the area of a polygon with the help of all the vertices of the polygon.

Formula: A = ½ (x1y2 - y1x2) + (x2y3 - y2x3) + …. + (xny1 - ynx1)

3. Give a brief overview of the distance from the origin.

• Distance from Origin:

Here, in NCERT Solutions for Class 10 Maths Chapter 7, you will have to find the distance from the origin to the given point, and the given point is also the origin.

• Section Formula:

In  Coordinate Geometry of NCERT Solutions for Class 10th Maths Chapter 7, you will be given points on the line segment and that line segment will be divided equally in a particular ratio and you will have to find the coordinate points.

• Midpoint Formula:

Here in Coordinate Geometry of NCERT Solutions for Class 10 Maths, you will be learning how to find the coordinate points of a given midpoint of the line segment.

4. How do I increase my ranks with Vedantu’s study guide?

You can improve your grades with our help. We at Vedantu have a team of experts who understand the academic needs and requirements of students. The study material provided by Vedantu is crafted as per the latest syllabus and is made available to everybody through our online portal.

Our solutions and study guide are designed by our subject-matter experts in a step by step way for easy and better understanding of the concept. We made sure that CBSE and NCERT guidelines are strictly followed while drafting these solutions. All the topics are explained in detail to make it more clear for you.

5. How many questions are there in Exercise 7.1 of Chapter 7 of Class 10 Maths?

Exercise 7.1 of Chapter 7 of Class 10 Maths consists of 10 questions in total. These questions are very well explained in detailed steps in NCERT solutions from Vedantu which will help the student to learn the concept of the chapter with full understanding. The practice from the solutions will give them confidence. Complex problems are made easy so that the students will not be demotivated.

6. What is Coordinate Geometry according to Chapter 7 of Class 10?

Coordinate geometry may be defined as a study of geometry with the help of a coordinate system. It is a section of geometry using which the positions of given points on a plane can be defined by a particular pair of numbers referred to as coordinates. Coordinate geometry is an important chapter and students have to learn it properly by paying full attention in class and also practise the chapter as much as possible.

These solutions are available on Vedantu's official website(vedantu.com) and mobile app free of cost.

7. What is the Distance Formula?

The distance formula in the chapter is used to find the distance between the two points in the plane of XY. The distance of the point from the Y-axis is called the X-coordinate and the distance from the X-axis is called the Y-coordinate. The point on the X-axis is(x,0) and the point in the Y-axis is(0, y). To find the distance on the plane the theorem of Pythagoras is used in an important chapter in Class 10.

The distance formula is PQ =√[(x2 – x1)2 + (y2 – y1)2]

Distance = √[(x2 – x1)2 + (y2 – y1)2

8. What are Collinear Points?

Collinear points refer to three or more than three points lying on a line. So, when they join and extend, it forms one straight line. In case the points aren't collinear, they will make a triangle when joined together. This word is derived from a Latin word Col+Linear. Col means together and linear means a straight line. These concepts have to be clear to solve any type of numericals connected with the chapter.

9. What are the different formulas for finding Collinear Points?

The three different formulas are the Distance formula, Slope formula and Area Of A Triangle.If A, B and C are the collinear points then AB+BC=AC

We know the distance between the point (x1,y1)and (x2,y2)

Distance =√(X2-X1)2+(Y2--Y1)2

The slope formula is used to measure the steepness of the line.

m=(X2-X1)/(Y2--Y1)

For triangle area can be found by

(1/2) | [x1(y2 – y3) + x2(y3 – y1) + x3[y1 – y2]| = 0

10. What is covered in Class 10 Chapter 7 Exercise 7.1 of Maths: Coordinate Geometry?

Class 10 Maths Exercise 7.1 Solutions focuses on the basics of coordinate geometry, including understanding the Cartesian plane, plotting points, and using the distance formula to find the distance between two points.

11. What are the four quadrants of the Cartesian plane in exercise 7.1 class 10th?

Following are the points for class 10 ch 7 maths ex 7.1:

• Quadrant I: Both x and y coordinates are positive (top-right).

• Quadrant II: x is negative, y is positive (top-left).

• Quadrant III: Both x and y coordinates are negative (bottom-left).

• Quadrant IV: x is positive, y is negative (bottom-right).

12. What are some common mistakes to avoid when solving problems in Class 10th maths chapter 7 exercise 7.1?

Coordinate Geometry Class 10 Exercise 7.1  gives:

• Incorrectly plotting points: Ensure accuracy when locating points on the Cartesian plane.

• Sign errors in the distance formula: Pay close attention to the signs of the coordinates when subtracting.

• Arithmetic mistakes: Double-check calculations, especially when squaring numbers and taking square roots.

13. Why is the distance formula important in class 10 ch 7 maths ex 7.1?

The distance formula is crucial in coordinate geometry as it allows you to determine the distance between any two points in the plane, which is fundamental for solving various geometric problems and understanding the relationships between points.