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

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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

If you are looking  for the comprehensive NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.2, then you are at the right place. Detailed NCERT Solutions for all sums covered in exercise 7.2 of NCERT Class 10 Maths are provided here in this PDF. The subject experts at Vedantu have prepared these solutions to help you prepare for the Class 10 Maths examination.

Similar to the solved examples, the exercise questions will help you in gaining knowledge of different types of sums. In order to score good marks in CBSE Class 10 board examination, students are suggested to study and practice  these exercise solutions as many times as possible.

Topics Covered in Exercise 7.2

  1. Section formula

  2. Solved examples

What is Section Formula?

The Section formula in coordinate geometry is used to determine the ratio in which a line segment is splitted by a point, either  internally or externally.

If a point K (lying on AB) divides AB in the ratio p:q, then section formula is given:
$\frac{\text{px2 +qx1}}{\text{p+q}},\frac{\text{py2 + qy1}}{\text{p+q}}$

Remember, the ratio p:q can also be written as p/q:1 of k:1, where k is equal to p/q.


Free PDF download of NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.2 (Ex 7.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise NCERT solution in your emails.


Class:

NCERT Solutions for Class 10

Subject:

Class 10 Maths

Chapter Name:

Chapter 7 - Coordinate Geometry

Exercise:

Exercise - 7.2

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes


You can also download Maths NCERT Solutions Class 10 to help you to revise complete syllabus and score more marks in your examination. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution Class 10 Science, Maths solutions and Solutions of other subjects.

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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.


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


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: 


the coordinates of the points of trisection of the line segment


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?


A rectangular shaped school ground ABCD


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\[{\text{(8,20)}}\]


Preet posted red flag at of the distance Q


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:


The ratio in which the line segment joining the points


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:


Coordinates of the point of division


 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: 


Vertices of a parallelogram


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: 


The coordinates of a point A, where AB is the diameter of circle whose centre is\[{\text{(2, - 3)}}\] and B is\[{\text{(1,4)}}\]


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


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:


The coordinates of the points which divide the line segment joining A\[{\text{( - 2,2)}}\] and B\[{\text{(2,8)}}\] into four equal parts


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: 


The area of a rhombus


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 = \[\dfrac{{\text{1}}}{{\text{2}}} \times{\text{ (product}}\;{\text{of}}\;{\text{lengths}}\;{\text{of}}\;{\text{diagonals)}}\]

                                      = \[\dfrac{{\text{1}}}{{\text{2}}}{\text{ AC}} \times {\text{BD}}\]

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


NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2

Opting for the NCERT solutions for Ex 7.2 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.2 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 7 Exercise 7.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 10 Maths Chapter 7 Exercise 7.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

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NCERT Solutions for Class 10 Maths Chapter 7 All Other Exercises

FAQs on NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry - Exercise 7.2

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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. Why should I refer to NCERT Solutions for Class 10 Maths Chapter 7?

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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 

For more clarification of your concepts, you can visit our website or the Vedantu app. 

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