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

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

Edited By Ramraj Saini | Updated on Nov 27, 2023 08:26 AM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.2

NCERT Solutions for Exercise 7.2 Class 10 Maths Chapter 7 Coordinate Geometry are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 7.2 deals with the concept of section formula which is used in mathematics to obtain the coordinates of the point that splits a line segment into a ratio either externally or internally. We can use the section formula to discover the coordinates of a point that divides a line segment externally or internally in some ratio. When a point P(x,y) divides the line segment into two segments, labelled as A (x1, y1) and B (x2,y2) then the method used to find the coordinates of that point is known as the section formula, which is mathematically represented as P(x, y) = [(mx2 + nx1)/(m + n), (my2 + ny1)/m + n]

NCERT solutions for exercise 7.2 Class 10 Maths chapter 7 Coordinate Geometry focus on section formula that divided into two sections namely the internal section formula and the external section formula. 10th class Maths exercise 7.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Assess NCERT Solutions for Class 10 Maths chapter 7 exercise 7.2

Coordinate Geometry Class 10 Chapter 7 Exercise: 7.2

Let the coordinates of point $P(x,y)$ which divides the line segment joining the points $A(-1,7)$ and $B(4,-3)$ , internally, in the ratio $m_{1}:m_{2}$ then,

Section formula: $\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Substituting the values in the formula:

Here, $m_{1}:m_{2} = 2:3$

$\Rightarrow \left (\frac{2(4)+3(-1)}{2+3} , \frac{2(-3)+3(7)}{2+3} \right )$

$\Rightarrow \left (\frac{5}{5} , \frac{15}{5} \right )$

Hence the coordinate is $P \left (1 , 3 \right )$ .

Let the trisection of the line segment $A(4,-1)$ and $B(-2,-3)$ have the points $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$

Then,

Section formula: $\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

By observation point, P divides AB internally in the ratio $1:2$ .

Hence, $m:n = 1:2$

Substituting the values in the equation we get;

$\Rightarrow P\left (\frac{1(-2)+2(4)}{1+2} , \frac{1(-3)+2(-1)}{1+2} \right )$

$\Rightarrow P \left (\frac{-2+8}{3} , \frac{-3-2}{3} \right )$

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

And by observation point Q, divides AB internally in the ratio $2:1$

Hence, $m:n = 2:1$

Substituting the values in the equation above, we get

$\Rightarrow Q\left (\frac{2(-2)+1(4)}{2+1} , \frac{2(-3)+1(-1)}{2+1} \right )$

$\Rightarrow Q \left (\frac{-4+4}{3} , \frac{-6-1}{3} \right )$

$\Rightarrow Q\left (0 , \frac{-7}{3} \right )$

Hence, the points of trisections are $\dpi{100} P \left (2 , \frac{-5}{3} \right )$ and $\dpi{100} Q\left (0 , \frac{-7}{3} \right )$

Niharika posted the green flag at the distance P, i.e.,

$\frac{1}{4}\times100\ m = 25\ m$ from the starting point of $2^{nd}$ line.

Therefore, the coordinates of this point $P$ are $(2,25).$

Similarly, Preet posted red flag at $\frac{1}{5}$ of the distance Q i.e.,

$\frac{1}{5}\times100\ m = 20\ m$ from the starting point of $8^{th}$ line.

Therefore, the coordinates of this point Q are $(8,20)$ .

The distance $PQ$ is given by,

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

and the point at which Rashmi should post her Blue Flag is the mid-point of the line joining these points. Let this point be $R(x,y)$ .

Then, by Section Formula,

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$x = \frac{2+8}{2},\ y = \frac{25+20}{2}$

$x = 5,\ y = 22.5$

Therefore, Rashmi should post her Blue Flag at 22.5 m on the 5th line.

Let the ratio be : $k:1$

Then, By section formula:

$P(x,y) = \left (\frac{kx_{2}+x_{1}}{k+1} , \frac{ky_{2}+y_{1}}{k+1} \right )$

Given point $P(x,y) = (-1,6)$

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

$\Rightarrow -k-1 = 6k-3$

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

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

Let the point on the x-axis be $P(x,0)$ and it divides it in the ratio $k:1$ .

Then, we have

Section formula:

$P(x,y) = \left (\frac{kx_{2}+x_{1}}{k+1} , \frac{ky_{2}+y_{1}}{k+1} \right )$

$\implies \frac{ky_{2}+y_{1}}{k+1} = 0$

$k =-\frac{y_{1}}{y_{2}}$

Hence, the value of k will be: $k =-\frac{-5}{5}= 1$

Therefore, the x-axis divides the line in the ratio $1:1$ and the point will be,

Putting the value of $k=1$ in section formula.

$P(x,0) = \left ( \frac{x_{2}+x_{1}}{2}, 0 \right )$

$P(x,0) = \left ( \frac{1-4}{2}, 0 \right ) = \left ( \frac{-3}{2}, 0 \right )$

Let the given points $A(1,2),\ B(4,y),\ C(x,6),\ D(3,5)$ .

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.

The coordinates of the point O when it is mid-point of AC.

$\left ( \frac{1+x}{2}, \frac{2+6}{2} \right ) \Rightarrow \left ( \frac{x+1}{2}, 4 \right )$

The coordinates of the point O when it is mid-point of BD.

$\left ( \frac{4+3}{2}, \frac{5+y}{2} \right ) \Rightarrow \left ( \frac{7}{2}, \frac{5+y}{2} \right )$

Since both coordinates are of same point O.

Therefore,

$\frac{x+1}{2} =\frac{7}{2}$ and $4 = \frac{5+y}{2}$

Or,

$x = 6\ and\ y = 3$

As the centre point $C(2,-3)$ will be the mid-point of the diameter AB.

Then, the coordinates of point A will be $A(x,y)$ .

Given point $B(1,4)$ .

Therefore,

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

$\frac{x+1}{2} = 2\ and\ \frac{y+4}{2} = -3$

$\Rightarrow x = 3\ and\ y = -10$ .

Therefore, the coordinates of A are $(3,-10).$

From the figure:

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

$\Rightarrow PB = \frac{4}{7}AB$ hence the ratio is 3:4,

Now, from the section formula, we can find the coordinates of Point P.

Section Formula:

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

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

$P(x,y)= \left (\frac{6-8}{7} , \frac{-12-8}{7} \right )$

$P(x,y)= \left (\frac{-2}{7} , \frac{-20}{7} \right )$

From the figure:

Points C, D, and E divide the line segment AB into four equal parts.

Now, from the section formula, we can find the coordinates of Point C, D, and E.

Section Formula:

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Here point D divides the line segment AB into two equal parts hence

$D(x_{2},y_{2})= \left (\frac{-2+2}{2} , \frac{2+8}{2} \right )$

$D(x_{2},y_{2})= \left (0 , 5 \right )$

Now, point C divides the line segment AD into two equal parts hence

$C(x_{1},y_{1})= \left (\frac{-2+0}{2} , \frac{2+5}{2} \right )$

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

Also, point E divides the line segment DB into two equal parts hence

$E(x_{1},y_{1})= \left (\frac{2+0}{2} , \frac{8+5}{2} \right )$

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

From the figure:

Let the vertices of the rhombus are:

$A(3,0),\ B(4,5),\ C(-1,4),\ D(-2,-1)$

Area of the rhombus ABCD is given by;

$= \frac{1}{2}\times(Product\ of\ lengths\ of\ diagonals)$

Hence we have to find the lengths of the diagonals AC and BD of the rhombus.

The distance formula:

$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of the diagonal AC:

$AC = \sqrt{(3-(-1))^2+(0-4)^2} = \sqrt{16+16} = 4\sqrt{2}$

Length of the diagonal BD:

$BD = \sqrt{(4-(-2))^2+(5-(-1))^2} = \sqrt{36+36} = 6\sqrt{2}$

Thus, the area will be,

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

$= \frac{1}{2}\times (4\sqrt{2})\times(6\sqrt{2}) = 24\ square\ units.$

## More About NCERT Solutions for Class 10 Maths Exercise 7.2: Section formula

Exercise 7.2 Class 10 Maths consists of a question based on the section formula. Internal section formula is used when a point P(x, y) divides the line segment AB internally into two segments in the ratio m:n which is expressed as P(x, y) = [(mx2 + nx1)/(m + n),(my2 + ny1)/(m + n)]. Similarly, the external section formula is used when a point P(x, y) divides the line segment AB externally into two segments in the ratio m:n which is expressed as P(x, y) = [(mx2 - nx1)/(m - n) ,(my2 - ny1)/(m - n)]. Section formula is also used to find the midpoint of a line segment. If M is the midpoint that divides the line segment AB in the ratio 1:1, then the coordinate of M is expressed as M(x, y) = [(x2 + x1)/2, (y2 + y1)/2] . The NCERT solutions for Class 10 Maths exercise 7.2 also focused on the distance formula. A few questions related to the distance formula are given also in exercise 7.2 Class 10 Maths. Students can find Coordinate Geometry Class 10 Notes here which can be used to quick revision of the concepts discussed in this chapter.

## Benefits of NCERT Solutions for Class 10 Maths Exercise 7.2 :

• NCERT solutions for Class 10 Maths exercise 7.2 helps us to understand how to find the coordinates of a point on a line segment AB which is divided in the ratio of m:n.
• The section formula can also be used to locate the midpoint of a line segment in exercise 7.2 Class 10 Maths.
• Students can achieve a good mark in term exams as well as competitive tests such as the joint admission exam by solving the NCERT solution for Class 10 Maths chapter 7 exercise 7.2 exercises.

Also see-

## Subject Wise NCERT Exemplar Solutions

1. What are the two subsections of the section formula?

Solution :

• Internal section formula

• External section formula

2. The section formula is the formula which is used for finding ________

The section formula is the formula that is used for finding the midpoint of a line segment.

3. What is m in the section formula?

A point P(x,y) divides the line segment AB into two segments in the ratio m:n. m is the ratio in dividing the line segment.

4. In what case is the section formula used to determine the coordinates of the point?

When a point on the line segment divides into two segments, then the section formula is used to determine the coordinates of that point.

5. What is the section formula consistent with NCERT solutions for Class 10 Maths chapter 7 exercise 7.2 ?

When some extent P(x,y) divides the segment into two segments, with marked points as A (x1,y1) and B(x2,y2) then the formula which is employed to work out the coordinates of that time is understood because of the section formula.

6. How many questions are there within the NCERT solutions for Class 10 Maths chapter 7 exercise 7.2 ?

There are 10 questions within the NCERT solutions for Class 10 Maths chapter 7 exercise 7.2 .

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

Yes, you can definitely apply for diploma courses after passing 10th CBSE. In fact, there are many diploma programs designed specifically for students who have completed their 10th grade.

Generally, passing 10th CBSE with a minimum percentage (often 50%) is the basic eligibility for diploma courses. Some institutes might have specific subject requirements depending on the diploma specialization.

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