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NCERT Solutions for Class 10 Chapter 7 Coordinate Geometry

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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry - Free PDF Download

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry provides you all the basic concepts of Coordinate Geometry Class 10. All the solutions are created by expert teachers at Vedantu. The overall concepts explained in these solutions are based on the CBSE syllabus. The untiring works of our experts have brought easy and understandable solutions for the problems related to Coordinate Geometry. We provide you with comprehensive solutions in a proper stepwise format. It will help you to score your best in the Mathematics Exam. Download Coordinate Geometry Class 10  NCERT Solutions from Vedantu.com. Also, you can revise and solve the important questions for class 10 Maths Exam 2024-25, using the updated CBSE solutions provided by us.

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Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry - Free PDF Download
2. Glance of NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry | Vedantu
3. Access Exercise Wise NCERT Solutions for Chapter 7 Maths Class 10
4. Exercises Under NCERT Solutions for Class 10 Coordinate Geometry
5. Table of All Formulas of Coordinate Geometry
6. Class 10 Maths Chapter 7: Exercises Breakdown
7. Other Study Material for CBSE Class 10 Maths Chapter 7
8. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

Glance of NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry | Vedantu

  • In this article we will learn what Is Coordinate Geometry and the importance of Coordinate Geometry in Maths.

  • Learn how to calculate distance between two given points by using distance formula, $\sqrt{[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]}$

  • Find coordinates of the point dividing the line with some ratio by using section formulas.

  • The coordinates of P will be $\left( \dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n} \right)$

  • Calculate area of triangle in terms of coordinates of its vertices by using formula Area of a Triangle $= \dfrac{1}{2} \left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|$

  • This article contains chapter notes, formula, exercises link and important questions for chapter 7 - Coordinate Geometry

  • There are two exercises (20 fully solved questions) in Class 10th Maths Chapter 7 Coordinate Geometry.


Access Exercise Wise NCERT Solutions for Chapter 7 Maths Class 10

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NCERT Solutions for Class 10 Chapter 7 Coordinate Geometry
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Exercises Under NCERT Solutions for Class 10 Coordinate Geometry

NCERT Solutions for Chapter 7 Class 10 Maths, "Coordinate Geometry" covers the following four exercises:


Exercise 7.1 - This exercise contains 10 questions, students are introduced to the concept of the Cartesian Plane, plotting points on the plane, and finding the coordinates of a point. Students learn how to identify the quadrants of the plane and the axes (x-axis and y-axis). The exercise contains questions related to finding the coordinates of a point, plotting a point on the plane, and identifying the quadrant in which a point lies.


Exercise 7.2 - This exercise contains 10 questions, students learn to find the distance between two points on a Cartesian Plane using the distance formula. They are also taught how to apply the midpoint formula to find the midpoint of a line segment. This exercise contains questions related to finding the distance between two points, finding the midpoint of a line segment, and applying the distance and midpoint formulas to solve problems.


Access NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry Exercise 7.1

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

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

Ans:

Given that,

Let the points be $\left( 2,3 \right)$ and $\left( 4,1 \right)$

To find the distance between the points $\left( 2,3 \right),\left( 4,1 \right)$.

Distance between two points is given by the Distance formula $=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

Here, ${{x}_{1}}=2$

 ${{x}_{2}}=4$

${{y}_{1}}=3$

${{y}_{2}}=1$  

Thus, the distance between $\left( 2,3 \right)$ and $\left( 4,1 \right)$ is given by,

$d=\sqrt{{{\left( 2-4 \right)}^{2}}+{{\left( 3-1 \right)}^{2}}}$

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

$=\sqrt{4+4}$

$=\sqrt{8}$

$=2\sqrt{2}$

$\therefore $The distance between $\left( 2,3 \right)$ and $\left( 4,1 \right)$ is $2\sqrt{2}$ units.


(ii) $\left( -5,7 \right),\left( -1,3 \right)$

Ans:

Given that,

Let the points be $\left( -5,7 \right)$ and $\left( -1,3 \right)$

To find the distance between the points $\left( -5,7 \right),\left( -1,3 \right)$.

Distance between two points is given by the Distance formula $=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

Here, ${{x}_{1}}=-5$

${{x}_{2}}=-1$

${{y}_{1}}=7$

${{y}_{2}}=3$  

Thus, the distance between $\left( -5,7 \right)$ and $\left( -1,3 \right)$ is given by,

$d=\sqrt{{{\left( -5-\left( -1 \right) \right)}^{2}}+{{\left( 7-3 \right)}^{2}}}$

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

$=\sqrt{16+16}$

$=\sqrt{32}$

$=4\sqrt{2}$

$\therefore $The distance between $\left( -5,7 \right)$ and $\left( -1,3 \right)$ is $4\sqrt{2}$ units.


(iii) $\left( a,b \right),\left( -a,-b \right)$

Ans:

Given that,

Let the points be $\left( a,b \right)$ and $\left( -a,-b \right)$

To find the distance between the points $\left( a,b \right),\left( -a,-b \right)$.

Distance between two points is given by the Distance formula $=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

Here, ${{x}_{1}}=a$

${{x}_{2}}=-a$

${{y}_{1}}=b$

${{y}_{2}}=-b$  

Thus, the distance between $\left( a,b \right)$ and $\left( -a,-b \right)$ is given by,

$d=\sqrt{{{\left( a-\left( -a \right) \right)}^{2}}+{{\left( b-\left( -b \right) \right)}^{2}}}$

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

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

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

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

$\therefore $The distance between $\left( a,b \right)$ and $\left( -a,-b \right)$ is $2\sqrt{{{a}^{2}}+{{b}^{2}}}$ units.


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

Ans:

Given that,

Let the points be $\left( 0,0 \right)$ and $\left( 36,15 \right)$

To find the distance between the points $\left( 0,0 \right),\left( 36,15 \right)$.

Distance between two points is given by the Distance formula $=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

Here, ${{x}_{1}}=0$

${{x}_{2}}=36$

${{y}_{1}}=0$

${{y}_{2}}=15$  

Thus, the distance between $\left( 0,0 \right)$ and $\left( 36,15 \right)$ is given by,

$d=\sqrt{{{\left( 0-36 \right)}^{2}}+{{\left( 0-15 \right)}^{2}}}$

$=\sqrt{{{\left( -36 \right)}^{2}}+{{\left( -15 \right)}^{2}}}$

$=\sqrt{1296+225}$

$=\sqrt{1521}$

$=39$

Yes, it is possible to find the distance between the given towns $A$ and $B$. The positions of this town are $A\left( 0,0 \right)$ and $B\left( 36,15 \right)$. And it can be calculated as above.

$\therefore $The distance between $A\left( 0,0 \right)$ and $B\left( 36,15 \right)$ is $39$ km.


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

Ans:

Given that,

Let the three points be $\left( 1,5 \right),\left( 2,3 \right)$ and $\left( -2,-11 \right)$

To determine if the given points are collinear

Let $A\left( 1,5 \right),B\left( 2,3 \right),C\left( -2,-11 \right)$ be the vertices of the given triangle.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $A\left( 1,5 \right)$ and $B\left( 2,3 \right)$

${{x}_{1}}=1$

${{x}_{2}}=2$

${{y}_{1}}=5$

${{y}_{2}}=3$

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

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

$=\sqrt{1+4}$

$=\sqrt{5}$

To find the distance between the points $B\left( 2,3 \right)$ and $C\left( -2,-11 \right)$

${{x}_{1}}=2$

${{x}_{2}}=-2$

${{y}_{1}}=3$

${{y}_{2}}=-11$

$BC=\sqrt{{{\left( 2-\left( -2 \right) \right)}^{2}}+{{\left( 3-\left( -11 \right) \right)}^{2}}}$

$=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 14 \right)}^{2}}}$

$=\sqrt{16+196}$

$=\sqrt{212}$

To find the distance between the points  $A\left( 1,5 \right)$  and  $C\left( -2,-11 \right)$

${{x}_{1}}=1$

${{x}_{2}}=-2$

${{y}_{1}}=5$

${{y}_{2}}=-11$

$CA=\sqrt{{{\left( 1-\left( -2 \right) \right)}^{2}}+{{\left( 5-\left( -11 \right) \right)}^{2}}}$

$=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 16 \right)}^{2}}}$

$=\sqrt{9+256}$

$=\sqrt{265}$

Since $AB+AC\ne BC$ and $AB\ne BC+AC$ 

and $AC\ne BC$

$\therefore $The points $A\left( 1,5 \right),B\left( 2,3 \right),C\left( -2,-11 \right)$ are not collinear.


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

Ans:

Given that,

Let the three points be $\left( 5,-2 \right),\left( 6,4 \right)$ and $\left( 7,-2 \right)$ are the vertices of the triangle.

To determine if the given points are the vertices of an isosceles triangle.

Let $A\left( 5,-2 \right),B\left( 6,4 \right),C\left( 7,-2 \right)$ be the vertices of the given triangle.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $A\left( 5,-2 \right)$ and $B\left( 6,4 \right)$

${{x}_{1}}=5$

${{x}_{2}}=6$

${{y}_{1}}=-2$

${{y}_{2}}=4$

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

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

$=\sqrt{1+36}$

$=\sqrt{37}$

To find the distance between the points $B\left( 6,4 \right)$ and $C\left( 7,-2 \right)$

${{x}_{1}}=6$

${{x}_{2}}=7$

${{y}_{1}}=4$

${{y}_{2}}=-2$

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

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

$=\sqrt{1+36}$

$=\sqrt{37}$

To find the distance between the points  $A\left( 5,-2 \right)$  and  $C\left( 7,-2 \right)$

${{x}_{1}}=5$

${{x}_{2}}=7$

${{y}_{1}}=-2$

${{y}_{2}}=-2$

$CA=\sqrt{{{\left( 5-7 \right)}^{2}}+{{\left( -2-\left( -2 \right) \right)}^{2}}}$

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

$=\sqrt{4+0}$

$=2$

We can conclude that $AB=BC$. 

Since two sides of the triangle are equal in length, $ABC$ is an isosceles triangle.


5. In a classroom, $4$ friends are seated at the points $A,B,C$ and $D$ are 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 the distance formula, find which of them is correct.

Ans:
Given that,

$4$ friends are seated at the points $A,B,C,D$

To find,

If they form square together by using distance formula


seo images


From the figure, we observe the points $A\left( 3,4 \right),B\left( 6,7 \right),C\left( 9,4 \right)$ and $D\left( 6,1 \right)$ are the positions of the four students.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $A\left( 3,4 \right)$ and $B\left( 6,7 \right)$

${{x}_{1}}=3$

${{x}_{2}}=6$

${{y}_{1}}=4$

${{y}_{2}}=7$

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

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

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$

To find the distance between the points $B\left( 6,7 \right)$ and $C\left( 9,4 \right)$

${{x}_{1}}=6$

${{x}_{2}}=9$

${{y}_{1}}=7$

${{y}_{2}}=4$

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

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

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$

To find the distance between the points  $C\left( 9,4 \right)$ and $\left( 6,1 \right)$

${{x}_{1}}=9$

${{x}_{2}}=6$

${{y}_{1}}=4$

$=2\left( lb+bh+lh \right)$

\[\]

$=2\left( 80 \right)CD=\sqrt{{{\left( 9-6 \right)}^{2}}+{{\left( 4-1 \right)}^{2}}}$

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

$=\sqrt{18}$

$=3\sqrt{2}$

To find the distance between the points $A\left( 3,4 \right)$ and $D\left( 6,1 \right)$

${{x}_{1}}=3$

${{x}_{2}}=6$

${{y}_{1}}=4$

${{y}_{2}}=1$

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

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

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$

Since all sides of the squares are equal, now find the distance between the diagonals $AC$ and $BD$.

To find the distance between the points $A\left( 3,4 \right)$ and $C\left( 9,4 \right)$

$=13\text{ cm}{{x}_{1}}=3$

${{x}_{2}}=9$

${{y}_{1}}=4$

${{y}_{2}}=4$

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

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

$=\sqrt{36+0}$

$=6$

Diagonal To find the distance between the points $B\left( 6,7 \right)$ and $D\left( 6,1 \right)$

${{x}_{1}}=6$

${{x}_{2}}=6$

${{y}_{1}}=7$

${{y}_{2}}=1$

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

$=\sqrt{{{\left( 0 \right)}^{2}}+{{\left( 6 \right)}^{2}}}$

$=\sqrt{0+36}$

$=6$

Thus, the four sides \[\] and $DA$ are equal and its diagonals $AC$ and $BD$ are also equal.

$\therefore ABCD$ form a square and hence Champa was correct.


6. Name the type of quadrilateral forms, if any, by the following points, and give reasons for your answer.

(i) $\left( -1,-2 \right),\left( 1,0 \right),\left( -1,2 \right),\left( -3,0 \right)$-

Ans:

Given that,

Let the given points denote the vertices $A\left( -1,-2 \right),B\left( 1,0 \right),C\left( -1,2 \right),D\left( -3,0 \right)$ denote the vertices of the quadrilateral.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $A\left( -1,-2 \right)$ and $B\left( 1,0 \right)$

${{x}_{1}}=-1$

${{x}_{2}}=1$

${{y}_{1}}=-2$

${{y}_{2}}=0$

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

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

$=\sqrt{4+4}$

$=3.5\text{ cm}=\sqrt{8}$

$=2\sqrt{2}$

To find the distance between the points $B\left( 1,0 \right)$ and $C\left( -1,2 \right)$

${{x}_{1}}=1$

${{x}_{2}}=-1$

${{y}_{1}}=0$

${{y}_{2}}=2$

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

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

$=\sqrt{4+4}$

$=\sqrt{8}$

$=2\sqrt{2}$

To find the distance between the points  $C\left( -1,2 \right)$ and \[\]

$=137.5{{x}_{1}}=-1$

${{x}_{2}}=-3$

${{y}_{1}}=2$

${{y}_{2}}=0$

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

$=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 2 \right)}^{2}}}$

$=\sqrt{4+4}$

$=\sqrt{8}$

$=2\sqrt{2}$

To find the distance between the points $D\left( -3,0 \right)$ and $A\left( -1,-2 \right)$ 

${{x}_{1}}=-3$

${{x}_{2}}=-1$

${{y}_{1}}=0$

${{y}_{2}}=-2$

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

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

$=\sqrt{4+4}$

$=\sqrt{8}$

$=2\sqrt{2}$

To find the diagonals of the given quadrilateral

To find the distance between the points $A\left( -1,-2 \right)$ and $C\left( -1,2 \right)$

${{x}_{1}}=-1$

${{x}_{2}}=-1$

$=l{{y}_{1}}=-2$

${{y}_{2}}=2$

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

$=\sqrt{{{\left( 0 \right)}^{2}}+{{\left( -4 \right)}^{2}}}$

$=\sqrt{0+16}$

$=4$

To find the distance between the points $B\left( 1,0 \right)$ and $D\left( -3,0 \right)$

${{x}_{1}}=1$

${{x}_{2}}=-3$

${{y}_{1}}=0$

${{y}_{2}}=0$

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

$=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 0 \right)}^{2}}}$

$=\sqrt{16+0}$

$=4$

Since all sides of the given quadrilateral are of the same measure and the diagonals are also the same length. 

$\therefore $The given points of the quadrilateral form a square.

(ii) $\left( -3,5 \right),\left( 3,1 \right),\left( 0,3 \right),\left( -1,-4 \right)$-

Ans:

Given that,

Let the given points denote the vertices $A\left( -3,5 \right),B\left( 3,1 \right),C\left( 0,3 \right),D\left( -1,-4 \right)$ denote the vertices of the quadrilateral.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $A\left( -3,5 \right)$ and $B\left( 3,1 \right)$

${{x}_{1}}=-3$

${{x}_{2}}=3$

${{y}_{1}}=5$

${{y}_{2}}=1$

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

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

$=\sqrt{36+16}$

$=\sqrt{52}$

$=2\sqrt{13}$

To find the distance between the points \[\] and $C\left( 0,3 \right)$

${{x}_{1}}=3$

${{x}_{2}}=0$

${{y}_{1}}=1$

${{y}_{2}}=3$

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

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

$=\sqrt{9+4}$

$=\sqrt{13}=14-5$

To find the distance between the points  $C\left( 0,3 \right)=45\pi $ and $\therefore D\left( -1,-4 \right)$

${{x}_{1}}=0$

${{x}_{2}}=-1$

${{y}_{1}}=3$

${{y}_{2}}=-4$

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

$=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 7 \right)}^{2}}}$

$=\sqrt{1+49}$

$=\sqrt{50}$

$=5\sqrt{2}$

To find the distance between the points $A\left( -3,5 \right)$ and $D\left( -1,-4 \right)$ 

${{x}_{1}}=-344\text{ }{{\text{m}}^{2}}$

${{x}_{2}}=-1$

${{y}_{1}}=5$

${{y}_{2}}=-4$

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

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

$=\sqrt{4+81}$

$=\sqrt{85}$

From the distance we found that,

$AB\ne BC\ne CD\ne AD$

By plotting the graph, we get,


seo images


From the graph above, we can note that the points $ABC$ are collinear.  

$\therefore $The quadrilateral cannot be formed by using the above points.

(iii) $\left( 4,5 \right),\left( 7,6 \right),\left( 4,3 \right),\left( 1,2 \right)$-

Ans:

Given that,

Let the given points denote the vertices $A\left( 4,5 \right),B\left( 7,6 \right),C\left( 4,3 \right),D\left( 1,2 \right)$ denote the vertices of the quadrilateral.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $A\left( 4,5 \right)$ and $B\left( 7,6 \right)$

${{x}_{1}}=4$

${{x}_{2}}=7$

${{y}_{1}}=5$

${{y}_{2}}=6$

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

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

$=\sqrt{9+1}$

$=\sqrt{10}$

To find the distance between the points $B\left( 7,6 \right)$ and $C\left( 4,3 \right)$

${{x}_{1}}=7$

${{x}_{2}}=4$

${{y}_{1}}=6$

${{y}_{2}}=3$

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

$=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 3 \right)}^{2}}}$

$=\frac{1.4}{2}=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$

To find the distance between the points $C\left(4,3 \right)$ and $D\left( 1,2 \right)$

${{x}_{1}}=4$

${{x}_{2}}=1$

${{y}_{1}}=3$

${{y}_{2}}=2$

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

$=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 1 \right)}^{2}}}$

$=\sqrt{9+1}$

$=\sqrt{10}$

To find the distance between the points x$A\left( 4,5 \right)$ and $D\left( 1,2 \right)$ 

${{x}_{1}}=4$

${{x}_{2}}=1$

${{y}_{1}}=5$

${{y}_{2}}=2$

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

$=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 3 \right)}^{2}}}$

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$

To find the diagonals of the given quadrilateral

To find the distance between the points $A\left( 4,5 \right)$ and $C\left( 4,3 \right)$

${{x}_{1}}=4$

${{x}_{2}}=4$

${{y}_{1}}=5$

${{y}_{2}}=3$

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

$=\sqrt{{{\left( 0 \right)}^{2}}+{{\left( 2 \right)}^{2}}}$

$=\sqrt{0+4}$

$=2$

To find the distance between the points $B\left( 7,6 \right)$ and $D\left( 1,2 \right)$

${{x}_{1}}=7$

${{x}_{2}}=1$

${{y}_{1}}=6$

${{y}_{2}}=2$

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

$=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( 4 \right)}^{2}}}$

$=\sqrt{36+16}$

$=\sqrt{52}$

$=2\sqrt{13}$

From the above calculation, the opposite sides of the quadrilateral are of same length and the diagonals are not of same length

$\therefore $The given points of the quadrilateral form a parallelogram.


Question 7:

Find the point on the $x$-axis which is equidistant from $\left( 2,-5 \right)$ and $\left( -2,9 \right)$.

Ans:

Given that,

$\left( 2,-5 \right)$

$\left( -2,9 \right)$

To find,

The point that is equidistant from the points $\left( 2,-5 \right)$ and $\left( -2,9 \right)$

Let us consider the points as $A\left( 2,-5 \right)$ and $B\left( -2,9 \right)$ and to find the equidistant point $P$.

Since the point is on $x$-axis, the coordinates of the required point is of the form $P\left( x,0 \right)$.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $P\left( x,0 \right)$ and $A\left( 2,-5 \right)$ 

${{x}_{1}}=x$

${{x}_{2}}=2$

${{y}_{1}}=0$

${{y}_{2}}=-5$

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

$=\sqrt{{{\left( x-2 \right)}^{2}}+25}$

To find the distance between the points $A\left( x,0 \right)$ and $B\left( -2,9 \right)$

${{x}_{1}}=x$

${{x}_{2}}=-2$

${{y}_{1}}=0$

${{y}_{2}}=9$

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

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

$=\sqrt{{{\left( x+2 \right)}^{2}}+81}$

Since the distance are equal in measure,

$PA=PB$

$\sqrt{{{\left( x-2 \right)}^{2}}+25}$ $=\sqrt{{{\left( x+2 \right)}^{2}}+81}$

Taking square on both sides,

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

By solving, we get,

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

$8x=-56$

$x=-7$

The coordinate is $\left( -7,0 \right)$.

$\therefore $The point that is equidistant from $\left( 2,-5 \right)$ and $\left( -2,9 \right)$ is $\left( -7,0 \right)$.


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

Ans:

Given that,

$P\left( 2,-3 \right)$

$Q\left( 10,y \right)$

The distance between these points is $10$ units.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

.To find the distance between the points $P\left( 2,-3 \right)$ and $Q\left( 10,y \right)$

${{x}_{1}}=2$

${{x}_{2}}=10$

${{y}_{1}}=-3$

${{y}_{2}}=y$

$PQ=\sqrt{{{\left( 10-2 \right)}^{2}}+{{\left( y-\left( -3 \right) \right)}^{2}}}$

$=\sqrt{{{\left( -8 \right)}^{2}}+{{\left( y+3 \right)}^{2}}}$

$=\sqrt{64+{{\left( y+3 \right)}^{2}}}$

Since the distance between them is $10$ units,

$\sqrt{64+{{\left( y+3 \right)}^{2}}}=10$

Squaring on both sides, we get,

$64+{{\left( y+3 \right)}^{2}}=100$

${{\left( y+3 \right)}^{2}}=36$

$y+3=\pm 6$

So, $y+3=6$

$y=3$

And, $y+3=-6$

$y=-9$

$\therefore $The possible values of $y$ are $y=3$ or $y=-9$.


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

Ans:

Given that,

$Q\left( 0,1 \right)$

$P\left( 5,-3 \right)$

$R\left( x,6 \right)$

To find,

  • The values of $x$

  • The distance of $QR$ and $PR$

$Q$ is equidistant between $P$ and $R$.

SO $PQ=QR$

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

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

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

$\sqrt{25+16}=\sqrt{{{x}^{2}}+25}$

Squaring on both sides, we get,

$41={{x}^{2}}+25$

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

 x = ±4

Thus the point $R$ is $R\left( 4,6 \right)$ or $\left( -4,6 \right)$.

To find the distance $PR$ and $QR$

Case (1):

When the point is $R\left( 4,6 \right)$

Distance between the point $P\left( 5,-3 \right)$ and $R\left( 4,6 \right)$ is,

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

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

$=\sqrt{1+81}$

$=\sqrt{82}$

Distance between the point $Q\left( 0,1 \right)$ and $R\left( 4,6 \right)$,

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

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

$=\sqrt{16+25}$

$=\sqrt{41}$

Case (2):

When the point is $R\left( -4,6 \right)$

Distance between the point $P\left( 5,-3 \right)$ and $R\left( 4,6 \right)$,

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

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

$=\sqrt{81+81}$

$=\sqrt{162}$

$=9\sqrt{2}$

Distance between the point $Q\left( 0,1 \right)$ and $R\left( 4,6 \right)$,

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

$=\sqrt{{{\left( 4 \right)}^{3}}+{{\left( -5 \right)}^{2}}}$

$=\sqrt{16+25}$

$=\sqrt{41}$


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

Ans:

Given that,

$\left( x,y \right)$ is equidistant from $\left( 3,6 \right)$ and $\left( -3,4 \right)$

To find,

The value of $\left( x,y \right)$

Let $P\left( x,y \right)$ is equidistant from $A\left( 3,6 \right)$ and $B\left( -3,4 \right)$

Since they are equidistant,

$PA=PB$

$=66\text{ c}{{\text{m}}^{3}}\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-6 \right)}^{2}}}=\sqrt{\left( x-{{\left( -3 \right)}^{2}} \right)+{{\left( y-4 \right)}^{2}}}$

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

Squaring on both sides, we get,

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

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

$36-16=12x+4y$

$3x+y=5$

$3x+y-5=0$

$\therefore $The relation between $x$ and $y$ is given by $3x+y-5=0$


Exercise (7.2)

1.Find the coordinates of the point which divides the join of $\left( -1,7 \right)$ and $\left( 4,-3 \right)$ in the ratio $2:3$

Ans:

Given that,

The points $A\left( -1,7 \right)$ and $B\left( 4,-3 \right)$

Ratio $m:n=$ $2:3$

To find,

The coordinates

Let $P\left( x,y \right)$ be the required coordinate


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By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$P\left( x,y \right)=\left[ \frac{2\left( 4 \right)+3\left( -1 \right)}{2+3},\frac{2\left( -3 \right)+3\left( 7 \right)}{2+3} \right]$

$=\left[ \frac{8-3}{5},\frac{-6+21}{5} \right]$

$=\left[ \frac{5}{5},\frac{15}{5} \right]$

$=\left( 1,3 \right)$

$\therefore $The coordinates of $P$ is $P\left( 1,3 \right)$.


2. Find the coordinates of the point of trisection of the line segment joining $\left( 4,-1 \right)$ and $\left( -2,-3 \right)$.

Ans:

Given that,

The line segment joining $\left( 4,-1 \right)$ and $\left( -2,-3 \right)$

To find,

The coordinates of the point 

Let the line segment joining the points be $A\left( 4,-1 \right)$ and $B\left( -2,-3 \right)$

Let $P\left( {{x}_{1}},{{y}_{1}} \right)$ and $Q\left( {{x}_{2}},{{y}_{2}} \right)$ are the points of trisection of the line segment joining the points 

$AP=PQ=QB$


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From the diagram, the point $P$ divides $AB$ internally in the ratio of $1:2$

Hence $m:n=1:2$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$P\left( {{x}_{1}},{{y}_{1}} \right)=\left[ \frac{1\left( -2 \right)+1\left( 4 \right)}{1+2},\frac{1\left( -3 \right)+2\left( -1 \right)}{1+2} \right]$

$=\left[ \frac{-2+8}{3},\frac{-3-2}{3} \right]$

$=\left[ \frac{6}{3},\frac{-5}{3} \right]$

$=\left( 2,-\frac{5}{3} \right)$

$\therefore $The coordinates of $P$ is $P\left( 2,-\frac{5}{3} \right)$.

From the diagram, the point $Q$ divides $AB$ internally in the ratio of $2:1$

Hence $m:n=2:1$

By section formula,

$Q\left( {{x}_{2}},{{y}_{2}} \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$Q\left( {{x}_{2}},{{y}_{2}} \right)=\left[ \frac{2\left( -2 \right)+1\left( 4 \right)}{2+1},\frac{2\left( -3 \right)+3\left( -1 \right)}{2+1} \right]$

$=\left[ \frac{-4+4}{3},\frac{-6-1}{3} \right]$

$=\left[ 0,-\frac{7}{3} \right]$

$\therefore $The coordinates of $Q$ is $Q\left( 0,-\frac{7}{3} \right)$.


3. To conduct Sports day activities, in your rectangular shaped school ground $ABCD$, lines have been drawn with chalk powder at a distance of $1\text{ m}$ each. $100$ flower pots have been placed at a distance of $1\text{ m}$ from each other along $AD$, as shown in the following figure. Niharika runs $\frac{1}{4}\text{ th}$ the distance $AD$ on the $2\text{nd}$ line and posts a green flag. Preet runs $\frac{1}{5}\text{ th}$the distance $AD$ on the eighth line and posts a red flag. What is the distance between both the flags?If Rashmi has to post a blue flag exactly halfway between the line segments joining the two flags, where should she post her flag?


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Ans:

Given that,

  • Niharika posted her green flag at a distance of $P$which is $\frac{1}{4}\times 100=25\text{ m}$from the starting point of the second line. The coordinate of $P$ is $P\left( 2,25 \right)$

  • Preet posted red flag at $\frac{1}{5}$ of a distance $Q$ which is $\frac{1}{5}\times 100=20\text{ m}$from the starting point of the eighth line. The coordinate of $Q$ is $\left( 8,20 \right)$


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The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

The distance between the flags $P\left( 2,25 \right),Q\left( 8,20 \right)$ is given by,

$PQ=\sqrt{{{\left( 8-2 \right)}^{2}}+{{\left( 25-20 \right)}^{2}}}$

$=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( 5 \right)}^{2}}}$

$=\sqrt{36+25}$

$=\sqrt{61}\text{ m}$

Rashmi should post her blue flag in the mid-point of the line joining points $P$ and $Q$. Let this point be $M\left( x,y \right)$.

By section formula,

$M\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$M\left( x,y \right)=\left[ \frac{1\left( 2 \right)+1\left( 8 \right)}{1+1},\frac{1\left( 25 \right)+1\left( 20 \right)}{1+1} \right]$

$=\left[ \frac{2+8}{2},\frac{25+20}{2} \right]$

$=\left[ \frac{10}{2},\frac{45}{2} \right]$

$=\left( 5,22.5 \right)$

$\therefore $Rashmi should place her blue flag at $22.5\text{ m}$ in the fifth line.


4. Find the ratio in which the line segment joining the points $\left( -3,10 \right)$ and $\left( 6,-8 \right)$ is divided by $\left( -1,6 \right)$

Ans:

Given that,

The line segment $\left( -3,10 \right)$ and $\left( 6,-8 \right)$

The point $\left( -1,6 \right)$ divides the line segment

To find,

The ratio of dividing line segment 


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Let the line segment be $A\left( -3,10 \right)$ and $B\left( 6,-8 \right)$ divided by point $P\left( -1,6 \right)$in the ratio of $k:1$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

Equating the $x$term,

$-1=\frac{6k-3}{k+1}$

$-k-1=6k-3$

$7k=2$

$k=\frac{2}{7}$

$\therefore $The point $P$ divides the line segment $AB$ in the ratio of $2:7$


5. Find the ratio in which the line segment joining $A\left( 1,-5 \right)$ and $B\left( -4,5 \right)$ is divided by the $x$ axis. Also find the coordinates of the point of division.

Ans:

Given that,

The line segment joining the points $A\left( 1,-5 \right)$ and $B\left( -4,5 \right)$

To find,

  • The ratio

  • The coordinates of the point of division


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Let the ratio be $k:1$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$P\left( x,y \right)=\left[ \frac{k\left( -4 \right)+1\left( 1 \right)}{k+1},\frac{k\left( 5 \right)+1\left( -5 \right)}{k+1} \right]$

$=\left[ \frac{-4k+1}{k+1},\frac{5k-5}{k+1} \right]$

We know that $y$ coordinate on $x$ axis is zero.

$\frac{5k-5}{k+1}=0$

$5k-5=0$

$5k=5$

$k=1$

Therefore $x$ axis divides it in the ratio of $1:1$

Division point, $P=\left( \frac{-4\left( 1 \right)+1}{1+1},\frac{5\left( 1 \right)-5}{1+1} \right)$

$=\left( \frac{-4+1}{2},\frac{5-5}{2} \right)$

$=\left( -\frac{3}{2},0 \right)$

$\therefore $The ratio at which the line segment is divided is $1:1$ and the point of division is  $\left( -\frac{3}{2},0 \right)$.


6. If $\left( 1,2 \right),\left( 4,y \right),\left( x,6 \right)$ and $\left( 3,5 \right)$ are the vertices of the parallelogram taken in order, find $x$ and $y$.

Ans:

Given that,

The vertices of the parallelogram are $A\left( 1,2 \right),B\left( 4,y \right),C\left( x,6 \right),D\left( 3,5 \right)$

To find,

The value of $x$ and $y$

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  • The diagonals of the parallelogram bisect each other at $O$. 

  • Intersection point $O$ of diagonal $AC$ and $BD$divides these diagonals. So $O$ is the midpoint of $AC$ and $BD$.

If $O$ is the midpoint of $AC$,

$O=\left( \frac{1+x}{2},\frac{2+6}{2} \right)$

$=\left( \frac{1+x}{2},\frac{8}{2} \right)$

$=\left( \frac{1+x}{2},4 \right)$

If $O$ is the midpoint of $BD$,

$O=\left( \frac{4+3}{2},\frac{5+y}{2} \right)$

$=\left( \frac{7}{2},\frac{5+y}{2} \right)$

Equating the points of $O$,

$\frac{x+1}{2}=\frac{7}{2}$ and 

$4=\frac{5+y}{2}$

Finding $x$ term,

$\frac{x+1}{2}=\frac{7}{2}$

$x+1=7$

$x=6$

Finding $y$ term,

$4=\frac{5+y}{2}$

$5+y=8$

$y=3$

The value of $x$ and $y$ are $x=6$ and $y=3$


7. Find the coordinates of point $A$, where $AB$ is the diameter of circle whose center is $\left( 2,-3 \right)$ and $B$ is $\left( 1,4 \right)$

Ans:

Given that,

  • Center is $C\left( 2,-3 \right)$

  • The coordinate of $B$ is $B\left( 1,4 \right)$

To find,

The coordinate of $A$

Let the coordinate of $A$ be $A\left( x,y \right)$


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Midpoint of $AB$ is $C\left( 2,-3 \right)$and so,

$\left( 2,-3 \right)=\left( \frac{x+1}{2},\frac{y+4}{2} \right)$

Equating $x$ term,

$\frac{x+1}{2}=2$

$x+1=4$

$x=3$

Equating $y$ term,

$\frac{y+4}{2}=-3$

$y+4=-6$

$y=-10$

$\therefore $The coordinate of $A$ is $A\left( 3,-10 \right)$


8. If $A$ and $B$ are $\left( -2,-2 \right)$ and $\left( 2,-4 \right)$ respectively, find the coordinates of $P$ such that $AP=\frac{3}{7}AB$ and $P$ lies on the line segment $AB$.

Ans:

Given that,

  • The coordinates are $A\left( -2,-2 \right)$ and $B\left( 2,-4 \right)$

  • $AP=\frac{3}{7}AB$

To find, 

The coordinate of $P$


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$AP=\frac{3}{7}AB$

$\frac{AB}{AP}=\frac{7}{3}$

From the figure, $AB=AP+PB$

$\frac{AP+PB}{AP}=\frac{3+4}{3}$

$1+\frac{PB}{AP}=1+\frac{4}{3}$

$\frac{PB}{AP}=\frac{4}{3}$

$\therefore AP:PB=3:4$

Thus, the point $P\left( x,y \right)$ divides the line segment $AB$ in the ratio of $3:4$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$P\left( x,y \right)=\left[ \frac{3\left( 2 \right)+4\left( -2 \right)}{3+4},\frac{3\left( -4 \right)+4\left( -2 \right)}{3+4} \right]$

$=\left[ \frac{6-8}{7},\frac{-12-8}{7} \right]$

$=\left[ -\frac{2}{7},-\frac{20}{7} \right]$

$\therefore $The coordinates of $P$ is $P\left( -\frac{2}{7},-\frac{20}{7} \right)$.


7. Find the coordinates of the points which divide the line segment joining $A\left( -2,2 \right)$ and $B\left( 2,8 \right)$ into four equal parts.

Ans;

Given that,

The line segment $A\left( -2,2 \right)$ and $B\left( 2,8 \right)$

To find,

The coordinate that divides the line segment into four equal parts


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Form the figure, ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ be the points that divide the line segment $AB$ into four equal parts.

Point ${{P}_{1}}$ divides the line segment $AB$ in the ratio of $1:3$, so, 

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

${{P}_{1}}\left( x,y \right)=\left[ \frac{1\left( 2 \right)+3\left( -2 \right)}{1+3},\frac{1\left( 8 \right)+3\left( 2 \right)}{1+3} \right]$

$=\left[ \frac{-4}{4},\frac{14}{4} \right]$

$=\left( -1,\frac{7}{2} \right)$

Point ${{P}_{2}}$ divides the line segment $AB$ in the ratio of $1:1$, so, 

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

${{P}_{2}}\left( x,y \right)=\left[ \frac{1\left( 2 \right)+1\left( -2 \right)}{1+1},\frac{1\left( 8 \right)+1\left( 2 \right)}{1+1} \right]$

$=\left[ \frac{2+\left( -2 \right)}{2},\frac{2+8}{2} \right]$

$=\left( 0,5 \right)$

Point ${{P}_{3}}$ divides the line segment $AB$ in the ratio of $3:1$, so, 

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

${{P}_{3}}\left( x,y \right)=\left[ \frac{3\left( 2 \right)+1\left( -2 \right)}{3+1},\frac{3\left( 8 \right)+1\left( 2 \right)}{3+1} \right]$

$=\left[ \frac{6-2}{4},\frac{24+2}{4} \right]$

$=\left( 1,\frac{13}{2} \right)$

The coordinates that divide the line segment into four equal parts are ${{P}_{1}}\left( -1,\frac{7}{2} \right),{{P}_{2}}\left( 0,5 \right)$ and ${{P}_{3}}\left( 1,\frac{13}{2} \right)$


10. Find the area of the rhombus if its vertices are $\left( 3,0 \right),\left( 4,5 \right),\left( -1,4 \right)$ and $\left( -2,-1 \right)$ taken in order. (Hint: Area of a rhombus = $\frac{1}{2}$(Product of its diagonals))

Ans:

Given that,

The vertices of the rhombus are $A\left( 3,0 \right),B\left( 4,5 \right),C\left( -1,4 \right)$ and $D\left( -2,-1 \right)$

To find, 

The area of the rhombus


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The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

Distance between the diagonal $AC$ is given by,

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

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

$=\sqrt{16+16}$

$=\sqrt{32}$

$=4\sqrt{2}$

Distance between the diagonal $BD$ si given by,

$BD=\sqrt{{{\left( 4-\left( -2 \right) \right)}^{2}}+{{\left( 5-\left( -1 \right) \right)}^{2}}}$

$=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( 6 \right)}^{2}}}$

$=\sqrt{36+36}$

$=\sqrt{72}$

$=6\sqrt{2}$

Area of the rhombus = $\frac{1}{2}\times $(Products of lengths of diagonals

$=\frac{1}{2}\times AC\times BD$

$=\frac{1}{2}\times 4\sqrt{2}\times 6\sqrt{2}$

$=24$ square units


What is Coordinate Geometry

Coordinate Geometry is the branch of mathematics that helps us to exactly locate a given point with the help of an ordered pair of numbers. Coordinate geometry is the combination of geometry and algebra to solve the problems. It helps you find the distance between two points whose coordinates are given. You can also find the coordinates of the point which divides the line segment joining two given points in the given ratio.


Coordinate geometry is also said to be the study of graphs. Graphs are visual representations of our data. It can be in different forms like bar graphs, histograms, line graphs, etc.


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Terms Related to Coordinate Geometry

While studying coordinate geometry the students must be aware of some important terms related to coordinate geometry they are well explained in this chapter.

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From this figure, let us understand some important terms used in the coordinate formula of geometry.

 

Axes of Coordinates

In the above figure OX and OY are called X-axis and Y-axis respectively and both together are known as axes of coordinates.

 

Origin

The point of intersection of the axes is called the origin; it is O.

 

Abscissa

The distance of any point on the plane from the Y-axis is called the abscissa.

 

Ordinate

The distance of any point on the plane from the X-axis is said to be ordinate.

 

Coordinate of the Origin

It has zero distance from both the axes. Therefore the coordinates of the origin are (0, 0).

 

Quadrant

The axes divide the plane into four parts. These four parts are said to be quadrants.

A quadrant is ¼ th part of a plane divided by coordinate axes. The plane is called the coordinate plane or the XY-plane and the axes are called the coordinate axes.

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In the first quadrant, both the coordinates are positive.

In the second quadrant, the y-coordinate is positive and x-coordinate is negative.

In the third quadrant, both the coordinates are negative.

In the fourth quadrant, the y-coordinate is negative and the x-coordinate is positive.

 

Distance Formula

In chapter 7 Maths Class 10 we learn about distance formulas for finding distance between two points. Finding the distance between the two points by using formula when the two coordinates of the points are given.

Distance between two points O(x1,y1) and B(x2, y2) =OB = \[\sqrt{[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]}\]

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Section Formula

Section formula helps us in finding the coordinates of the point dividing the line in the ratio m:n.

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

The coordinates of P will be $\left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right)$

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Area of Triangle

It will let us find the area of any triangle in terms of coordinates of its vertices. This formula will also be used in finding the area of quadrilaterals.

Area of a Triangle = ½ |x1(y2−y3)+x2(y3–y1)+x3(y1–y2)|

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All the basic formulas of coordinate geometry will help you to solve your geometry and algebra problems. 

Chapter 7 maths Class 10 provides you the list of all formulas of coordinate geometry to solve all the related problems.


Table of All Formulas of Coordinate Geometry

General Form of a Line

Ax + By + C = 0

Where A , B, C are real numbers and x and y are variables

Slope Intercept Form of a Line

$y = m \times x + c$

Where x and y are variables, c is constant and m is the slope

Point-Slope Form

$Y-y1= m\times(x - x1)$

x1 ,y1,x2,y2 are the X Y coordinates and m is the slope

The slope of a Line Using Coordinates

$m = \dfrac{Δy}{Δx} = \dfrac{(y2 − y1)}{(x2 − x1)}$

The slope of a Line Using General Equation

$m = −(\dfrac{A}{B})$

Intercept-Intercept Form

$\dfrac{x}{a} +\dfrac{y}{b} = 1$

Distance between two points O(x1,y1) and B(x2, y2)

OB= \[\sqrt{[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]}\]

For Parallel Lines,

m1 = m2

For Perpendicular Lines,

$m1\times m2 = -1$

Midpoint Formula/Section Formula 

M (x, y) = [½(x1 + x2), ½(y1 + y2)]

Angle between Two Lines 

$θ = tan${-1} \left[\dfrac{(m_1 - m_2)}{ 1 + m_1m_2}\right]$

Area of a Triangle

½ |x1(y2−y3)+x2(y3–y1)+x3(y1–y2)|

Perpendicular Distance from a Point to a Line

$d = [\dfrac{|ax1 + by1 + c|}{ √(a2 + b2)}]$



Class 10 Maths Chapter 7: Exercises Breakdown

Chapter 7 Coordinate Geometry Exercises in PDF Format

Exercise 7.1

10 Questions & Solutions 

Exercise 7.2

10 Questions & Solutions


Conclusion

NCERT Solutions for Class 10 Chapter 7 Coordinate Geometry offers a thorough grasp of this crucial subject.  students can develop a strong foundation in coordinate geometry by concentrating on essential ideas such as recognising slopes, finding distances, and charting points. It's important to pay attention to the step-by-step solutions provided in the NCERT Solutions, as they help clarify concepts and reinforce problem-solving techniques. Coordinate geometry is important to comprehend because it provides the foundation for understanding more advanced mathematical ideas and has real-world applications in computer graphics, engineering, and architecture. In previous year question papers, around 5-6 questions have been typically asked from this chapter.


Other Study Material for CBSE Class 10 Maths Chapter 7



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 Chapter 7 Coordinate Geometry

1.  How to find the circumcentre of a triangle by the section formula?

Circumcentre is defined as the point of intersection of the perpendicular bisectors of the sides of a triangle.

The coordinates of circumcentre of triangle is calculated whose coordinates of vertices are given, by the application of section formula i.e., midpoint formula. The following steps are involved for calculating the coordinates of circumcentre:

Step1: Using the midpoint formula, find the coordinates of midpoint (xm, ym), of each side AB, BC and AC.

Step2: Calculate the slope of each side. If the slope of any side is ‘m1’ then, the slope of line perpendicular to it will be 1/m1. Assume, m = 1/m1.

Step3: Using coordinates of midpoint (xm,ym), and the slope of perpendicular line i.e., ‘m’. write the equation of line (y-ym) = m (x-xm).

Step4: Similarly, find the equation of other lines.

Step5: Solve them to find out their intersection point.

The obtained intersection point will be the circumcentre of the given triangle.

2.  What will be the equation of straight line having 2-unit distance from the origin and passing through the point (4, -2)?

Let the equation of straight line be given as ax + by + c = 0.

The perpendicular distance from any point is given by formula: | ax+by+c / √(a²+b²) |

Given that, the straight line is having 2-unit distance from origin (0,0). 

So, | a(0)+b(0)+c / √(a²+b²) | = 2

⇒ | c / √(a²+b²) | = 2

⇒ c = ± 2√(a²+b²)

Assume, value of c = - 2√(a²+b²) …..(1)

Also, the straight line passes through the point (4, -2). Putting the coordinates of given point in the equation of straight line, we get:

4a – 2b + c = 0 

Replacing the value of ‘c’ from equation (1), we get:

4a – 2b - 2√(a²+b²) = 0

⇒ 4a – 2b = 2√(a²+b²)

Square both sides of above equation,

(4a – 2b)2 = (2√(a²+b²)2

⇒ 16a2 + 4b2 – 16ab = 4(a²+b²) (cancel out ‘4b2’ on both sides)

⇒ 12a2 = 16ab (cancel out ‘a’ on both sides) 

⇒ 3a = 4b

⇒ b = 3a/4 …..(2)

Put the value of ‘b’ obtained in equation (1) to get the value ‘c’ in terms of ‘a’.

c = - 2√(a²+b²)

⇒ c = - 2√(a²+9(a)² / 16) = -2√[25(a)² / 16]

⇒ c = -(5/2 a) …..(3)

Now, put the values of ‘b’ and ‘c’ from equation (2) and equation (3) respectively in the assumed equation of straight line i.e., ax + by + c = 0.

ax + (3a/4)y - (5/2 a) = 0. (where, a ≠ 0, ‘a’ can be any arbitrary number except 0)

Let us assume the value of ‘a’ = 4. Then,

Required equation of straight line will be: 4x + 3y - 10 = 0

3. How is the area of a triangle formed by three points is expressed in Coordinate Geometry?

Let ABC be a triangle whose vertices are A (x1, y1), B (x2, y2) and C (x3, y3). 

Draw AP, BQ and CR perpendiculars from vertices A, B and C, respectively, to the x-axis. Clearly ABQP, APRC and BQRC are all trapezium.

From above figure, 

Area of Δ ABC = Area of trapezium ABQP + Area of trapezium APRC – Area of trapezium BQRC

We know that the area of trapezium = 1/2 (sum of parallel sides) (distance between them)

Therefore, area of Δ ABC = 1/2 (BQ + AP) QP + 1/2 (AP + CR) PR – 1/2 (BQ + CR) QR

= 1/2 (y2 + y1) (x1 – x2) + 1/2 (y1 + y3) (x3 – x1) – 1/2 (y2 + y3) (x3 – x2

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

Thus, the area of a triangle formed by three points in coordinate geometry is numerical value of expression: 1/2 |[x1(y2 -y3) + x2(y3 -y1) + x3(y1 – y2)]|.

4. What are some real-world applications for Coordinate Geometry?

Coordinate Geometry is used in navigation systems, computer graphics, and even designing buildings!

5. What are the concepts covered in ch7 Class 7 Maths Coordinate Geometry?

Ch7 Class 10 Maths coordinate geometry includes an introduction to coordinate geometry. Coordinate Geometry is the branch of mathematics that helps us to exactly locate a given point with the help of an ordered pair of numbers. Coordinate geometry is also known as cartesian geometry. Coordinate geometry class 10 covers the concepts of how to find the distance between two points whose coordinates are given. Find the coordinates of the point which divides the line segment joining two given points in the given ratio. Finally students will learn how to find the area of a triangle in terms of the coordinate of its vertices.

6. What is called a Coordinate Plane?

A two-dimensional plane that is formed by the intersection of one horizontal line called the x-axis and one vertical line called the y-axis. These lines are perpendicular to each other and intersect at a point O called the Origin. This axis divides the plane into four parts called quadrants. This plane is called the coordinate plane or XY- plane. The coordinate plane is also called the Cartesian plane. And the axis is called the coordinate axis.



7. Why do we need Coordinate Geometry?

Coordinate geometry has various applications in our day to daily life. Some of the areas where coordinate geometry is used are given as follow:

  • In computers, mobile phones, etc. to locate the position of cursor or any finger touch.

  • To determine the exact location of an airplane, trains etc while moving.

  • In GPS locations in digital maps to point the exact locations.

  • Coordinate geometry is also used in constructions of roads, buildings, etc.

  • Coordinate geometry is also used in visualizing 2D and 3D objects in the cartesian plane.

8. How do I plan my study for Class 10 exams?

To study and score well in Class 10 CBSE exams, students need to develop an effective strategy that will help them cover the syllabus. Here are some tips to plan your studies well in advance;

  • Go through the complete syllabus 

  • Go through the unit-wise weightage of each chapter. You can find it here.

  • Make a list of important formulas from each chapter

  • Make sure to practice well all the exercise questions

  • Revise the concepts well before the exams

9. What are the most important formulas that I need to remember in Chapter 7 Coordinate Geometry of Class 10 Maths?

Chapter 7 of  NCERT Class 10 Maths is Coordinate Geometry. Students should remember the following important formulas from this chapter -

  • The distance formula 

  • The section formula

  • Area of a triangle using coordinates

  • The midpoint of a line segment joining two points

  • Distance of a point from the origin

Students should practice the examples given in the chapter and the exercise questions at the end to understand the concepts clearly and to score well in exams.

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

10. Where can I find the solutions to Chapter 7 Coordinate Geometry of Class 10 Maths?

Students can find the solutions to all exercise questions of NCERT Class 10 Maths from Vedantu’s website. The solutions are curated by subject experts to make the studies easier for students :

  • Go to Vedantu.

  • Here you will see all the chapters listed on the page.

  • Click on the chapter for which you want the solutions.

  • The solutions will appear on the screen.

  • You can also download the solutions for all the chapters by clicking on ‘Download PDF’. These can be viewed offline too.

11. Are there any negative coordinates in Coordinate Geometry?

Yes! Points can lie to the left (negative x) or below (negative y) of the origin (0,0) on the coordinate plane.

12. How many exercises are there in Chapter 7 of Coordinate Geometry of Class 10 Maths?

There are a total of four exercises. To understand how to solve them easily, you can visit Vedantu to avail the NCERT Solutions curated by experts. That’s not all, right from important questions to revision notes, you will find several tools to help you become thorough in this chapter.