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# NCERT Solutions for Class 10 Maths 7 Exercise 7.2 - Coordinate Geometry

Last updated date: 24th Jul 2024
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## NCERT Solutions for 10 Maths 7 Coordinate Geometry Exercise 7.2 - FREE PDF Download

Chapter 7 of Ex 7.2 Class 10 Maths, called "Coordinate Geometry," helps you learn about geometry using a coordinate system, which is like a map with an x-axis and a y-axis. Exercise 7.2 specifically deals with finding the distance between two points, dividing a line segment into a certain ratio, and calculating the area of a triangle using coordinates. The NCERT Solutions for this exercise provide clear, step-by-step answers to the problems, making it easier for you to understand and solve them. These solutions will help you get better at solving geometry problems and prepare well for your exams.

Table of Content
1. NCERT Solutions for 10 Maths 7 Coordinate Geometry Exercise 7.2 - FREE PDF Download
2. Glance on NCERT Solutions for Class 10 Maths Chapter 7  Coordinate Geometry - Exercise 7.2
3. Topics Covered in Class 10 Maths Chapter 7 Exercise 7.2
4. Access NCERT Solutions for Class-10 Maths Chapter 7 – Coordinate Geometry
5. NCERT Solutions for Class 10 Maths Chapter 7 all Other Exercises
6. CBSE Class 10 Maths Chapter 7 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

## Glance on NCERT Solutions for Class 10 Maths Chapter 7  Coordinate Geometry - Exercise 7.2

• The primary objective is to explain the meaning of the Section Formula and how it works to calculate the ratio at which a point splits a line segment.

• The ratio by which a point divides a line segment, either internally or externally, can be found using the Section Formula in coordinate geometry.

• Types of division are described in this chapter, including finding out how a line segment is divided by a point both internally and externally.

• The midpoint formula is a specific version of the section formula used when the point divides the segment in a 1:1 ratio.

• For solving geometric shape problems (triangles, quadrilaterals, etc.) and locating centroids, incenters, and other important locations in geometric objects using the Section Formula.

• This chapter of Chapter 7 Maths Ex 7.2 Class 10 helps students understand coordinate geometry.

• There are links to video tutorials explaining Chapter 7 Ex 7.2 Class 10 Coordinate Geometry for better understanding.

• In Class 10th Maths, Chapter 7, Exercise 7.2 Coordinate Geometry there are 10 Solved Questions.

## Topics Covered in Class 10 Maths Chapter 7 Exercise 7.2

• Section Formula:

• Internal and External Division

• Mid-Point Formula

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## Access NCERT Solutions for Class-10 Maths Chapter 7 – Coordinate Geometry

Exercise-7.2

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

Ans: Let P${\text{(x,y)}}$ be the required point.

Let A${\text{( - 1,7)}}$ and B${\text{(4, - 3)}}$ where ${{\text{x}}_{\text{1}}}{\text{ = - 1,}}{{\text{y}}_{\text{1}}}{\text{ = 7,}}{{\text{x}}_{\text{2}}}{\text{ = 4,}}{{\text{y}}_{\text{2}}}{\text{ = - 3}}$ and ${\text{m:n = 2:3}}$

So, by section formula

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

Now, substituting the values we get,

${\text{x = }}\left[ {\dfrac{{{\text{2(4) + 3( - 1)}}}}{{{\text{2 + 3}}}}} \right]{\text{ = }}\dfrac{{{\text{8 - 3}}}}{{\text{5}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{5}}}{\text{ = 1}}$

${\text{y = }}\left[ {\dfrac{{{\text{2( - 3) + 3(7)}}}}{{{\text{2 + 3}}}}} \right]{\text{ = }}\dfrac{{{\text{ - 6 + 21}}}}{{\text{5}}}{\text{ = }}\dfrac{{{\text{15}}}}{{\text{5}}}{\text{ = 3}}$

Therefore, the co-ordinates of point P are${\text{(1,3)}}$

2. Find the coordinates of the points of trisection of the line segment joining${\text{(4, - 1)}}$ and${\text{( - 2, - 3)}}$

Ans:

Let line segment joining the points be A${\text{(4, - 1)}}$  and B${\text{( - 2, - 3)}}$.

Let P${\text{(}}{{\text{x}}_1}{\text{,}}{{\text{y}}_1}{\text{)}}$ and Q${\text{(}}{{\text{x}}_2}{\text{,}}{{\text{y}}_2}{\text{)}}$ are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

So, by section formula

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

Therefore, by observation point P divides AB internally in the ratio ${\text{1:2}}$

Now, substituting the values we get,

${{\text{x}}_1}{\text{ = }}\left[ {\dfrac{{{\text{1( - 2) + 2(4)}}}}{{1 + 2}}} \right]{\text{ = }}\dfrac{{{\text{ - 2 + 8}}}}{3}{\text{ = }}\dfrac{6}{3}{\text{ = 2}}$

${{\text{y}}_1}{\text{ = }}\left[ {\dfrac{{{\text{1( - 3) + 2( - 1)}}}}{{{\text{1 + 2}}}}} \right]{\text{ = }}\dfrac{{{\text{ - 3 - 2}}}}{3}{\text{ = }}\dfrac{{{\text{ - 5}}}}{3}$

Therefore, P${\text{(}}{{\text{x}}_1}{\text{,}}{{\text{y}}_1}{\text{)}}$=${\text{(2, - }}\dfrac{{\text{5}}}{{\text{3}}}{\text{)}}$

Therefore, by observation point Q divides AB internally in the ratio ${\text{2:1}}$

Now, substituting the values we get,

${{\text{x}}_1}{\text{ = }}\left[ {\dfrac{{{\text{2( - 2) + 1(4)}}}}{{2 + 1}}} \right]{\text{ = }}\dfrac{{{\text{ - 4 + 4}}}}{3}{\text{ = }}\dfrac{0}{3}{\text{ = 0}}$

${{\text{y}}_1}{\text{ = }}\left[ {\dfrac{{{\text{2( - 3) + 1( - 1)}}}}{{2 + 1}}} \right]{\text{ = }}\dfrac{{{\text{ - 6 - 1}}}}{3}{\text{ = }}\dfrac{{{\text{ - 7}}}}{3}$

Therefore, Q${\text{(}}{{\text{x}}_2}{\text{,}}{{\text{y}}_2}{\text{)}}$= ${\text{(0, - }}\dfrac{7}{{\text{3}}}{\text{)}}$

3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of ${\text{1 m}}$each. ${\text{100}}$ flower pots have been placed at a distance of${\text{1 m}}$from each other along AD, as shown in the following figure. Niharika runs$\dfrac{{\text{1}}}{{\text{4}}}{\text{th}}$ the distance AD on the 2nd line and posts a green flag. Preet runs$\dfrac{{\text{1}}}{5}{\text{th}}$ 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?

Ans: By observation, that Niharika posted the green flag at of the distance P i.e., ${\text{(}}\dfrac{{\text{1}}}{{\text{4}}}{ \times\text{ 100)}}\;{\text{m = 25}}\;{\text{m}}$  from the starting point of 2nd line. Therefore, the coordinates of this point P is${\text{(2,25)}}$

Similarly, Preet posted red flag at of the distance Q i.e., ${\text{(}}\dfrac{{\text{1}}}{5}{ \times \text{100)}}\;{\text{m = 20}}\;{\text{m}}$     from the starting point of 8th line. Therefore, the coordinates of this point Q are${\text{(8,20)}}$

We know that the distance between the two points is given by the distance formula

i.e. $\sqrt {{{{\text{(}}{{\text{x}}_{\text{1}}}{\text{ - }}{{\text{x}}_{\text{2}}}{\text{)}}}^{\text{2}}}{\text{ + (}}{{\text{y}}_{\text{1}}}{\text{ - }}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$

To find the distance between these flags PQ by substituting the values we get,

PQ =$\sqrt {{{{\text{(8 - 2)}}}^{\text{2}}}{\text{ + (25 - 20}}{{\text{)}}^{\text{2}}}} {\text{ = }}\sqrt {{\text{36 + 25}}} {\text{ = }}\sqrt {{\text{61}}} \;{\text{m}}$

The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be M${\text{(x,y)}}$

So, by section formula

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

Now, substituting the values we get,

${\text{x = }}\dfrac{{{\text{2 + 8}}}}{2}{\text{ = }}\dfrac{{10}}{2}{\text{ = 5}}$

${\text{y = }}\dfrac{{25 + 20}}{2}{\text{ = }}\dfrac{{{\text{45}}}}{2}{\text{ = 22}}{\text{.5}}$

Therefore, Rashmi should post her blue flag at ${\text{22}}{\text{.5 m}}$ on fifth line.

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

Ans:

Let the ratio in which the line segment joining A${\text{( - 3,10)}}$ and B${\text{(6, - 8)}}$ is divided by point P${\text{( - 1,6)}}$ be ${\text{k:1}}$.

So, by section formula

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

$\therefore \;{\text{ - 1 = }}\dfrac{{{\text{6k - 3}}}}{{{\text{k + 1}}}}$

${\text{ - k - 1 = 6k - 3}}$

${\text{7k = 2}}$                             (by cross multiplying and transposing)

${\text{k = }}\dfrac{2}{7}$

Hence the point P divides AB in the ratio$2:7$

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

Ans:

Let the ratio be ${\text{k:1}}$and let the line segment joining A$(1, - 5)$ and B$( - 4,5)$

So, by section formula

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

Now, by substituting the values we get the coordinate of the point of division ${\text{[}}\dfrac{{{\text{ - 4k + 1}}}}{{{\text{k + 1}}}}{\text{,}}\dfrac{{{\text{5k - 5}}}}{{{\text{k + 1}}}}{\text{]}}$

We know that y-coordinate of any point on x-axis is ${\text{0}}$ $\therefore {\text{ }}\dfrac{{{\text{5k - 5}}}}{{{\text{k + 1}}}} = 0 \Rightarrow {\text{5k}} - 5 = 0 \Rightarrow {\text{5k}} = 5 \Rightarrow {\text{k}} = 1$

Therefore, x-axis divides it in the ratio $1:1$

Division point = ${\text{(}}\dfrac{{{\text{ - 4(1) + 1}}}}{{{\text{1 + 1}}}}{\text{,}}\dfrac{{{\text{5(1) + 5}}}}{{{\text{1 + 1}}}}{\text{) = (}}\dfrac{{{\text{ - 4 + 1}}}}{{\text{2}}}{\text{,}}\dfrac{{{\text{5 + 5}}}}{{\text{2}}}{\text{) = (}}\dfrac{{{\text{ - 3}}}}{{\text{2}}}{\text{,0)}}$

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

Ans:

Let A${\text{(1,2)}}$, B${\text{(4, y)}}$,C${\text{(x,6)}}$ and D${\text{(3,5)}}$ are the vertices of a parallelogram ABCD. Since the diagonals of a parallelogram bisect each other, Intersection point O of diagonal AC and BD also divides these diagonals Therefore, O is the mid-point of AC and BD.

If O is the mid-point of AC, then the coordinates of O are

${\text{(}}\dfrac{{{\text{1 + x}}}}{{\text{2}}}{\text{,}}\dfrac{{{\text{2 + 6}}}}{{\text{2}}}{\text{)}} \Rightarrow {\text{(}}\dfrac{{{\text{x + 1}}}}{{\text{2}}}{\text{,4)}}$

If O is the mid-point of BD, then the coordinates of O are${\text{(}}\dfrac{{{\text{4 + 3}}}}{{\text{2}}}{\text{,}}\dfrac{{{\text{5 + y}}}}{{\text{2}}}{\text{)}} \Rightarrow {\text{(}}\dfrac{{\text{7}}}{{\text{2}}}{\text{,}}\dfrac{{{\text{5 + y}}}}{{\text{2}}}{\text{)}}$

Since both the coordinates are of the same point O, $\dfrac{{{\text{1 + x}}}}{{\text{2}}}{\text{ = }}\dfrac{{\text{7}}}{{\text{2}}}\;{\text{and}}\;\dfrac{{{\text{5 + y}}}}{{\text{2}}}{\text{ = 4}}$

${\text{x + 1 = 7}}\;{\text{and}}\;{\text{5 + y = 8}}$

${\text{x = 6}}\;{\text{and}}\;{\text{y = 3}}$

7. Find the coordinates of a point A, where AB is the diameter of circle whose centre is${\text{(2, - 3)}}$ and B is${\text{(1,4)}}$

Ans:

Let the coordinates of point A be${\text{(x,y)}}$

Mid-point of AB is C${\text{(2, - 3)}}$, which is the center of the circle.

${\text{(2, - 3) = (}}\dfrac{{{\text{x + 1}}}}{{\text{2}}}{\text{,}}\dfrac{{{\text{y + 4}}}}{{\text{2}}}{\text{)}}$

$\dfrac{{{\text{x + 1}}}}{{\text{2}}}{\text{ = 2 and}}\;\dfrac{{{\text{y + 4}}}}{{\text{2}}} = - 3$

${\text{x + 1 = 4 and}}\;{\text{y + 4 = - 6 }}$

${\text{x = 3 and}}\;{\text{y = - 10}}$.

So, coordinates of A are ${\text{(3, - 10)}}$

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

Ans:

The coordinates of point A and B are${\text{( - 2, - 2)}}$ and${\text{(2, - 4)}}$ respectively and ${\text{AP = }}\dfrac{{\text{3}}}{{\text{7}}}{\text{AB}}$ so, $\dfrac{{{\text{AP}}}}{{{\text{AB}}}}{\text{ = }}\dfrac{{\text{3}}}{{\text{7}}}$

We know that AB = AP + PB from figure,

$\dfrac{{{\text{AP + PB}}}}{{{\text{AP}}}}{\text{ = }}\dfrac{{{\text{3 + 4}}}}{3}$

$1 + \dfrac{{{\text{PB}}}}{{{\text{AP}}}}{\text{ = 1 + }}\dfrac{{\text{4}}}{3}$

$\dfrac{{{\text{PB}}}}{{{\text{AP}}}}{\text{ = }}\dfrac{{\text{4}}}{3}$

Therefore, ${\text{PB:AP = 4:3}}$

Point P${\text{(x,y)}}$ divides the line segment AB in the ratio${\text{3:4}}$

Coordinates of P${\text{(x,y)}}$ = ${\text{(}}\dfrac{{{\text{3 * 2 + 4 * ( - 2)}}}}{{{\text{3 + 4}}}}{\text{,}}\dfrac{{{\text{3 * ( - 4) + 4 * ( - 2)}}}}{{{\text{3 + 4}}}}{\text{)}}$

= ${\text{(}}\dfrac{{{\text{6 - 8}}}}{{\text{7}}}{\text{,}}\dfrac{{{\text{ - 12 - 8}}}}{{\text{7}}}{\text{)}}\;{\text{ = }}\;{\text{(}}\dfrac{{{\text{ - 2}}}}{{\text{7}}}{\text{,}}\dfrac{{{\text{ - 20}}}}{{\text{7}}}{\text{)}}$

9. Find the coordinates of the points which divide the line segment joining A${\text{( - 2,2)}}$ and B${\text{(2,8)}}$ into four equal parts.

Ans:

By observation, ${{\text{P}}_1}$,${{\text{P}}_2}$,${{\text{P}}_3}$ that points divides the line segment A${\text{( - 2,2)}}$and B${\text{(2,8)}}$into four equal parts. Point${{\text{P}}_1}$ divides the line segment A${{\text{P}}_2}$ into two equal parts

Hence, Coordinates of ${{\text{P}}_1}$= $(\dfrac{{1 \times 2 + 3 \times ( - 2)}}{{1 + 3}},\dfrac{{1 \times 8 + 3 \times 2}}{{1 + 3}})\; = \;( - 1,\dfrac{7}{2})$

Point${{\text{P}}_2}$ divides the line segment AB into two equal parts

Coordinates of ${{\text{P}}_2}$= $(\dfrac{{2 + ( - 2)}}{2},\dfrac{{2 + 8}}{2})\; = \;(0,5)$

Point ${{\text{P}}_{\text{3}}}$ divides the line segment B ${{\text{P}}_2}$ into two equal parts

Coordinates of ${{\text{P}}_3}$= $(\dfrac{{3 \times 2 + 1 \times ( - 2)}}{{1 + 3}},\dfrac{{3 \times 8 + 1 \times 2}}{{1 + 3}})\; = \;(1,\dfrac{{13}}{2})$

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

Ans:

Let A${\text{(3,0)}}$,B${\text{(4,5)}}$,C${\text{( - 1,4)}}$ and D${\text{( - 2, - 1)}}$ are the vertices of a rhombus ABCD.

We know that the distance between the two points is given by the distance formula

i.e. $\sqrt {{{{\text{(}}{{\text{x}}_{\text{1}}}{\text{ - }}{{\text{x}}_{\text{2}}}{\text{)}}}^{\text{2}}}{\text{ + (}}{{\text{y}}_{\text{1}}}{\text{ - }}{{\text{y}}_{\text{2}}}{{\text{)}}^{\text{2}}}}$

Therefore, distance between A${\text{(3,0)}}$and C${\text{( - 1,4)}}$ is given by

Length of diagonal AC = $\sqrt {{{[3 - ( - 1)]}^2} + {{(0 - 4)}^2}} = \sqrt {16 + 16} = 4\sqrt 2$

Therefore, distance between B${\text{(4,5)}}$and D ${\text{( - 2, - 1)}}$ is given by

Length of diagonal BD = $\sqrt {{{[4 - ( - 2)]}^2} + {{(5 - ( - 1))}^2}} = \sqrt {36 + 36} = 6\sqrt 2$

Area of rhombus ABCD = $\dfrac{{\text{1}}}{{\text{2}}} \times{\text{ (product}}\;{\text{of}}\;{\text{lengths}}\;{\text{of}}\;{\text{diagonals)}}$

= $\dfrac{{\text{1}}}{{\text{2}}}{\text{ AC}} \times {\text{BD}}$

= $\dfrac{{\text{1}}}{{\text{2}}} \times{\text{ 4}}\sqrt {\text{2}} \times{\text{6}}\sqrt {\text{2}} {\text{ = 24}}\;{\text{square}}\;{\text{units}}$

## Conclusion

The NCERT Solutions for class 10 maths chapter 7 exercise 7.2 on Coordinate Geometry, provided by Vedantu, offer comprehensive explanations and step-by-step solutions to all problems in this exercise. This chapter is crucial as it deals with key concepts such as finding the distance between two points, determining the coordinates of a point dividing a line segment in a given ratio, and calculating the area of a triangle using coordinate geometry. Practising these problems will build a strong foundation in coordinate geometry, which is beneficial for higher studies as well.

## NCERT Solutions for Class 10 Maths Chapter 7 all Other Exercises

 Exercise Number of Questions Exercise 7.1 10 Questions & Solutions (3 Short Answers, 7 Long Answer)

## 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 7 Exercise 7.2 - Coordinate Geometry

1. Where should I find NCERT solutions for Exercise 7.2 of Class 10 Maths?

You can find all the solutions to your questions on the Vedantu website (vedantu.com) or the app. Vedantu offers NCERT solutions free of cost. These solutions are prepared by experts and are presented in a very clear and precise manner so that you can understand the concepts well.

2. Write the important questions from Exercise 7.2 of Class 10 Maths?

All questions are important from Exercise 7.2 of Class 10 Maths. Solutions of Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2 Questions help you to revise the complete syllabus and score more marks. Vedantu offers authentic information as they are revised and checked before publishing. You can also download Maths NCERT Solutions Class 10 free of cost to help you to revise the complete syllabus and score more marks in your examination.

3. What topics are covered in exercise 7.2 class 10 pdf of NCERT Class 10 Maths Chapter 7 ?

In Class 10 Ex 7.2 of NCERT  Maths Chapter 7, students delve into calculating how far apart two points are on a coordinate plane by using the Distance Formula. This exercise helps build a practical understanding of how to apply mathematical concepts to measure straight-line distances between any two given points, making it a fundamental skill in geometry.

4. How to find the distance between two points in Chapter 7 Maths of Class 10?

The distance formula, which is an application of the Pythagorean theorem, is used to calculate the distance between two locations. The straight line distance may be calculated using the horizontal and vertical distances between two locations as follows:

Distance between two points A(a,b) and B(c,d)= (a-c)2 + (b-d)2

5. How Vedantu helps me in preparing for Class 10 Maths Chapter 7?

The best thing about Vedantu NCERT Solutions is that it is easily accessible and explained in a concise manner. Vedantu provides NCERT Solutions for Class 10 Maths Chapter 7 free of cost. All the solutions are written by the best Maths teachers in India. All these solutions are based on the latest exam pattern, CBSE guidelines and marking schemes. Students can download these solutions and save on their computers and access them anytime they want to.

6. How many questions are there in exercise 7.2 class 10 ?

In Exercise 7.2, students are presented with 10 questions that test different aspects and applications of the Distance Formula. These questions range from straightforward calculations to more complex problems that require students to integrate different concepts of coordinate geometry, ensuring a comprehensive understanding of the topic. This variety helps build a solid foundation in geometric calculations and problem-solving skills.

7. What should students focus on in Ex 7.2 Class 10 ?

Focus on mastering the application of the Distance Formula accurately and understanding the derivation and its practical applications in different scenarios.

8. Can mastering exercise 7.2 class 10 help in competitive exams as well?

Mastering class 10 maths Ch 7 Ex 7.2 equips students with the ability to tackle a range of problems in competitive settings where geometric and spatial analysis is required. The skill to accurately compute distances and relate geometric properties using algebra is invaluable, not just in exams like the JEE, but also in problem-solving challenges across many science and engineering disciplines.