**NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry** – Here are all the NCERT solutions for Class 10 Maths Chapter 7. This solution contains questions, answers, images, explanations of the complete Chapter 7 titled Coordinate Geometry of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 7 Coordinate Geometry. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry in one place.

## NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Maths for Class 10 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 7 Coordinate Geometry , Maths, Class 10.

Class | 10 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 7 |

Chapter Name |
Coordinate Geometry |

### NCERT Solutions Class 10 Maths chapter 7 Coordinate Geometry

Class 10, Maths chapter 7, Coordinate Geometry solutions are given below in PDF format. You can view them online or download PDF file for future use.

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### Download NCERT Solutions Class 10 Maths chapter 7 Coordinate Geometry In PDF Format

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### Question & Answer

Q.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)

Ans :(i) distance between the two points is given by \( \sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\) Therefore. distance between (2, 3) and (4, 1) is given by \( l=\sqrt{(2-4)^{2}+(3-1)^{2}}=\sqrt{(-2)^{2}+(2)^{2}}\) \( =\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\) (ii) Distance between (-S,7) and (-1,3) is given by \( l=\sqrt{(-5-(-1))^{2}+(7-3)^{2}}=\sqrt{(-4)^{2}+(4)^{2}}\) \( =\sqrt{16+16}=\sqrt{32}=4 \sqrt{2}\) (iii) Distance between (a, b) and (-a -b) is given by \( l=\sqrt{(a-(-a))^{2}+(b-(-b))^{2}}\) \( =\sqrt{\left((2 a)^{2}+(2 b)^{2}\right.}=\sqrt{4 a^{2}+4 b^{2}}=2 \sqrt{a^{2}+b^{2}}\)

Q.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 discussed in Section 7.2.

Ans :Distance between points (0.0) and (36,15) \( =\sqrt{(36-0)^{2}+(15-0)^{2}}=\sqrt{36^{2}+15^{2}}\) \( =\sqrt{1296+225}=\sqrt{1521}=39\) Yes, we can find the distance between the given towns A and B Assume town A at origin point (O, O), Therefore, town a Rill be at point (36, 15) with respect to town A. And hence, as calculated above, the distance between town A and B Bill be 39 km

Q.3:Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Ans :Let the points (1, 5), (2, 3), and (-2, -11) be representing the vertices A, B, and C of the given triangle respectively. Let A=(1,5) , B =(2,3) , C=(-2,-11) \( \therefore A B=\sqrt{(1-2)^{2}+(5-3)^{2}}=\sqrt{5}\) \(BC=\sqrt{(2-(-2))^{2}+(3-(-11))^{2}}=\sqrt{4^{2}+14^{2}}=\sqrt{16+196}=\sqrt{212}\) \( CA=\sqrt{(1-(-2))^{2}+(5-(-11))^{2}}=\sqrt{3^{2}+16^{2}}=\sqrt{9+256}=\sqrt{265}\) Since \( \mathrm{AB}+\mathrm{BC} \neq \mathrm{CA}\) Therefore, the points (1, 5), (2, 3), and (-2, -11) are not collinear,

Q.4:Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Ans :Let the points (5, -2), (6, 4), and (7, -2) are representing the vertices A, B, and C of the given triangle respectively. \( A B=\sqrt{(5-6)^{2}+(-2-4)^{2}}=\sqrt{(-1)^{2}+(-6)^{2}}=\sqrt{1+36}=\sqrt{37}\) \( B C=\sqrt{(6-7)^{2}+(4-(-2))^{2}}=\sqrt{(-1)^{2}+(6)^{2}}=\sqrt{1+36}=\sqrt{37}\) \( \mathrm{CA}=\sqrt{(5-7)^{2}+(-2-(-2))^{2}}=\sqrt{(-2)^{2}+0^{2}}=2\) Therefore AB=BC As two sides are equal in length, therefore, ABC is an isosceles triangle.

Q.5:In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. 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 :\( A B=\sqrt{(3-6)^{2}+(4-7)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) \( B C=\sqrt{(6-9)^{2}+(7-4)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) \( C B=\sqrt{(9-6)^{2}+(4-1)^{2}}=\sqrt{(3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) \( A D=\sqrt{(3-6)^{2}+(4-1)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) \( A C=\sqrt{(3-9)^{2}+(4-4)^{2}}=\sqrt{(-6)^{2}+0^{2}}=6\) \( \mathrm{BD}=\sqrt{(6-6)^{2}+(7-1)^{2}}=\sqrt{0^{2}+(6)^{2}}=6\) It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are Of the sarne length, Therefore, ABCD is a square and hence, Champa correct

## NCERT / CBSE Book for Class 10 Maths

You can download the NCERT Book for Class 10 Maths in PDF format for free. Otherwise you can also buy it easily online.

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### All NCERT Solutions Class 10

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### All NCERT Solutions

You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

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**Exercise Covered under Maths Book Class 10 Maths Chapter 7 NCERT Solutions**

Here on this page we have covered following exercises from NCERT Maths Book for Class 10 Maths Chapter 7 Coordinate Geometry

- Exercise: 7.1 Class 10 Maths Solutions
- Exercise: 7.2 Class 10 Maths Solutions
- Exercise: 7.3 Class 10 Maths Solutions
- Exercise: 7.4 Class 10 Maths Solutions

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