**NCERT Solutions for Class 10 Maths Chapter 7 â€“ Coordinate Geometry**

NCERT Solutions for class 10 maths chapter 7 – Coordinate geometry will help you to make your foundation strong on the concepts of Coordinate geometry class 10. The study of Coordinate geometry class 10 Â and solving the problems will help you to solve complex problems easily. Coordinate geometry class 10 covers all the exercises provided in the NCERT textbook.

CBSE Class 10 Maths Chapter 7 NCERT Solutions are prepared by our expert at Toppr to help you to prepare for your exams in a better way and enhance your score. Coordinate geometry class 10 Â provide step by step solutions for the questions given in class 10 maths NCERT textbook as per CBSE Board guidelines and are also prepared according to the exam pattern. With the Toppr app, you can download NCERT Solutions for class 10 maths chapter 7 for free. In case you have a doubt while you are studying, Coordinate geometry class 10, for this we have a team of teachers who prove live doubt solving sessions only for you.

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**CBSE Class 10 Maths Chapter 7 NCERT SolutionsÂ **

Coordinate geometry class 10 explains the distance between the two points whose coordinates, area of the triangle formed by three given points, coordinates of the point which divides a line segment joining two points in a given ratio, Distance formula, Section formula, Area of a Triangle. Also, the questions are solved with alternative solutions and diagrammatic representation.

### Coordinate Geometry Sub-topics

- 7.1 – Introduction to Coordinate Geometry
- 7.2 – Distance Formula
- 7.3 – Section Formula
- 7.4 – Area of a Triangle
- 7.5 – Summary

**You can download NCERT Solutions for Class 10 Maths Chapter 7 PDF for free by clicking on the button below.**

**NCERT Solutions for Class 10 Maths Chapter 7**

**Q.1Â Find the distance between the following pairs of points:Â **

(i)Â

(2, 3), (4, 1)Â

(ii)Â

(âˆ’5, 7), (âˆ’1, 3)Â

(iii)Â

(a, b), (âˆ’a, âˆ’b)Â

**SolutionÂ **

Distance between two points ( x1 , y1 ) and ( x2 , y2Â ) isÂ

âˆš âˆ’ ( xâˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ 1 âˆ’ x2)2 + ( y1 âˆ’ y2)âˆ’ 2 (i) Distance between (2, 3) and (4, 1)Â

isÂ

= âˆš âˆ’ (2 âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ âˆ’ 4 )2 + (3 âˆ’ 1 )âˆ’ 2 = âˆš âˆ’ (âˆ’2 âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ (2)2 + )âˆ’ 2 = âˆš âˆ’ 4+4 âˆ’âˆ’âˆ’ = 8âˆš = 2 2âˆšÂ

(ii) Distance between (âˆ’5, 7) and (âˆ’1, 3)Â

isÂ

= âˆš âˆ’ (âˆ’5 âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ âˆ’ (âˆ’1) )2 + (7 âˆ’ 3 )âˆ’ 2 = âˆš âˆ’ (âˆ’4 âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ (4)2 + )âˆ’ 2 = âˆš âˆ’ 16 âˆ’âˆ’âˆ’âˆ’âˆ’ + 16 = âˆš âˆ’âˆ’ 32= 4 2âˆšÂ

(iii )Distance between (a, b) and (âˆ’a, âˆ’b)Â

isÂ

= âˆš âˆ’ (aâˆ’ âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ (âˆ’a) )2 + (b âˆ’ (âˆ’b) )âˆ’ 2 = âˆš âˆ’ (2a âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ )2 + (2b )âˆ’ 2 = âˆš âˆ’ 4 âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ a2 + 4 b2 = 2 âˆš âˆ’ aâˆ’âˆ’âˆ’âˆ’âˆ’ 2 + b2 #465323 Topic: Distance Between Two PointsÂ

**Q.2 Find the distance between the points (0, 0) and (36, 15) .Â **

**SolutionÂ **

Distance Between two given point=Â

âˆš âˆ’ ( xâˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ 2 âˆ’ x1)2 + ( y2 âˆ’ y1)âˆ’ 2 HereÂ

x1 = 0, x2 = 26 and y1 = 0, y2 = 15 âˆ´ Distance between the points (0, 0) and (36, 15) =Â

âˆš âˆ’ (36 âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ âˆ’ 0 )2 + (15 âˆ’ 0 )âˆ’ 2 = âˆš âˆ’ +36âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ 2 152 = âˆš âˆ’ 1296 âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ + 225 âˆ’ = âˆš âˆ’ 1521 âˆ’âˆ’âˆ’ = 39Â

**Q.3 Determine if the points (1, 5), (2, 3) and (âˆ’2, âˆ’11)Â are collinear.Â **

**SolutionÂ **

Three points A , B and CÂ

are collinear ifÂ

AB+BC = ACÂ

Here, point A(1, 5), B(2, 3) andÂ

C(âˆ’2, âˆ’11). âˆ´ AB = âˆš âˆ’ (2 âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ âˆ’ 1 )2 + (3 âˆ’ 5 )âˆ’ 2 = âˆš âˆ’ 1âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ 2 +(âˆ’ 22 âˆ’ ) = âˆš âˆ’ 1+4 âˆ’âˆ’âˆ’ = 5âˆš = 2.23 BC = âˆš âˆ’ ((âˆ’2) âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ âˆ’ (2) )2 + ((âˆ’11) âˆ’ (3) )âˆ’ 2 = âˆš âˆ’ (âˆ’4 âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ )2 + (âˆ’14 )âˆ’ 2 = âˆš âˆ’ 16 âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ + 196 = âˆš âˆ’âˆ’âˆ’ 212 = 14.56 AC = âˆš âˆ’ ((âˆ’2) âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ âˆ’ (1) )2 + ((âˆ’11) + (5) )âˆ’ 2 = (âˆš âˆ’ 3 )2 + (âˆ’16 )2 = âˆš âˆ’ 9 âˆ’âˆ’âˆ’âˆ’âˆ’ + 256 = âˆš âˆ’âˆ’âˆ’ 265 = 16.27 AB+BC = 2.23 + 14.56 = 16.79Â

AB+BC â‰ƒ ACÂ

Hence the given points are collinear.Â

**Q. 4Â Find the ratio in which the line segment joining the points (âˆ’3, 10) and (6, âˆ’8) is divided by (âˆ’1, 6)Â **

**SolutionÂ **

Let the required ratio beÂ

m1 : m2Â

Take ( x1 , y1 ) = (âˆ’3, 10); ( x2 , y2 ) = (6, âˆ’8) andÂ

(x,y) = (âˆ’1, 6) âˆ´ â‡’ x âˆ’1 = = m1xmm2 1 1 + m2x1Â

+ Ã—6+ m2Â

m2 Ã—âˆ’3 m1 + m2 â‡’âˆ’ m1 âˆ’ m2 =6m âˆ’ 3 m2Â

â‡’âˆ’ m1 âˆ’6 m1 =âˆ’3 m2 + m2Â

â‡’ âˆ’7 m1 = âˆ’2 m2 â‡’ m1m2Â

= 27 âˆ´ The required ratio is 2:7Â

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