NCERT Solutions for Exercise 7.2 Class 10 Maths Chapter 7

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

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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]

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NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.2
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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 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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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,

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

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

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-

NCERT Solutions Subject Wise

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Frequently Asked Questions (FAQs)

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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sai raja 01 September,2024

show the previous year exams ?

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as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

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i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

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i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

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

NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.2

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