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

Komal MiglaniUpdated on 30 Apr 2025, 03:13 PM IST

The section formula identifies its applications through the exercise of coordinate geometry. The exercise demonstrates how to calculate the point coordinates when dividing a line segment between two specified points by using a particular ratio. Students learn to find Cartesian plane positions when they solve these problems because such problems are essential foundations of basic geometric understanding. Knowledge about point positioning and their spatial relationships proves vital to tackle complex geometrical issues as well as identify certain spatial relations between points.

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  1. NCERT Solutions Class 10 Maths Chapter 7: Exercise 7.2
  2. Access Solution of Coordinate Geometry Class 10 Chapter 7 Exercise: 7.2
  3. Topics covered in Chapter 7 Coordinate Geometry: Exercise 7.2
  4. NCERT Solutions of Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject Wise
NCERT Solutions for Exercise 7.2 Class 10 Maths Chapter 7 - Coordinate Geometry
exercise 7.2

The section in the NCERT Solutions guide for Class 10 Maths demonstrates how to apply the section formula that appears in the NCERT textbooks. The instructional material presents students with problems about line segment division, both internally and externally, together with problems about calculating midpoints and applications to real-world examples. Learners who study Exercise 7.2 of the NCERT Books reinforce coordinate geometry understanding while developing essential analytical and problem-solving abilities for advanced mathematics.

NCERT Solutions Class 10 Maths Chapter 7: Exercise 7.2

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Access Solution of Coordinate Geometry Class 10 Chapter 7 Exercise: 7.2

Q1 Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Answer:

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 $P \left (1 , 3 \right )$ .

Q2 Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Answer:

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 $P \left (2 , \frac{-5}{3} \right )$ and $Q\left (0 , \frac{-7}{3} \right )$

Q3 To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th 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 segment joining the two flags, where should she post her flag

1638427152215

Answer:

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.

Q4 Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

Answer:

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 $2:7$ .

Q5 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.

Answer:

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.

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

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

Q6 If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Answer:

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,

$x = 6\ and\ y = 3$

Q7 Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).

Answer:

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 $(3,-10).$

Q8 If A and B are (- 2, - 2) and (2, - 4), respectively, find the coordinates of P such that AP = 3 AB 7 and P lies on the line segment AB.

Answer:

From the figure:

1638427203509

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:

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

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

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

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

Q9 Find the coordinates of the points which divide the line segment joining A (– 2, 2) and B(2, 8) into four equal parts.

Answer:

From the figure:

1638427222556

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

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

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

Q10 Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

Answer:

From the figure:

1638427243029

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,

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

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


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Topics covered in Chapter 7 Coordinate Geometry: Exercise 7.2

1. Section Formula: The section formula allows one to discover coordinates by dividing a straight line segment with a defined ratio.

2. Internal and External Division: Students must solve problems concerning line segment division when they involve internal and external divisions of these segments.

3. Mid-Point Formula: The section formula provides a method for determining the midpoint position between two points based on coordinate systems.

4. Real-Life Applications: The section formula finds real-world uses for practical problems that include determining positions in sports fields and construction layouts.

5. Analytical Geometry Skills: Using Analytical Geometry skills helps students improve their abilities to analyse geometrical problems by working with coordinates.

Check Out-

NCERT Solutions of Class 10 Subject Wise

Students must check the NCERT solutions for class 10 of the Mathematics and Science Subjects.

NCERT Exemplar Solutions of Class 10 Subject Wise

Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

Q: What are the two subsections of the section formula?
A:

Solution : 

  • Internal section formula

  • External section formula 

Q: The section formula is the formula which is used for finding ________
A:

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

Q: What is m in the section formula?
A:

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. 

Q: In what case is the section formula used to determine the coordinates of the point?
A:

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

Q: What is the section formula consistent with NCERT solutions for Class 10 Maths chapter 7 exercise 7.2 ?
A:

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.

Q: How many questions are there within the NCERT solutions for Class 10 Maths chapter 7 exercise 7.2 ?
A:

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

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