CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26
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.
This Story also Contains
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.
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 $Q\left (0 , \frac{-7}{3} \right )$
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.
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 $2:7$ .
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.
$P(x,0) = \left ( \frac{x_{2}+x_{1}}{2}, 0 \right )$
$P(x,0) = \left ( \frac{1-4}{2}, 0 \right ) = \left ( \frac{-3}{2}, 0 \right )$
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,
$x = 6\ and\ y = 3$
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 $(3,-10).$
Answer:
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:
$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
$P(x,y)= \left (\frac{3(2)+4(-2)}{3+4} , \frac{3(-4)+4(-2)}{3+4} \right )$
$P(x,y)= \left (\frac{6-8}{7} , \frac{-12-8}{7} \right )$
$P(x,y)= \left (\frac{-2}{7} , \frac{-20}{7} \right )$
Answer:
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
$E(x_{1},y_{1})= \left (\frac{2+0}{2} , \frac{8+5}{2} \right )$
$E(x_{2},y_{2})= \left (1 , \frac{13}{2} \right )$
Answer:
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,
$= \frac{1}{2}\times (AC)\times(BD)$
$= \frac{1}{2}\times (4\sqrt{2})\times(6\sqrt{2}) = 24\ square\ units.$
Also Read,
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-
Students must check the NCERT solutions for class 10 of the Mathematics and Science Subjects.
Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.
Frequently Asked Questions (FAQs)
Solution :
Internal section formula
External section formula
The section formula is the formula that is used for finding the midpoint of a line segment.
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.
When a point on the line segment divides into two segments, then the section formula is used to determine the coordinates of that point.
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.
There are 10 questions within the NCERT solutions for Class 10 Maths chapter 7 exercise 7.2 .
On Question asked by student community
Hello Rajesh!!!
I have question paper of science of 9th standard . Here is your question paper link where you have click to open and wrap it up.
https://school.careers360.com/boards/dge-tamil-nadu/tn-class-9-quarterly-exam-question-paper-2025-26
Hello,
For your information, the class 7 2025 result of the CM Shri School is expected to be published on September 20, 2025. So, if you want to see and download the results, then you can visit the official website of the CM Shri School: edudel.nic.in.
I hope it will clear your query!!
Hello
Visit the official website of the Rajasthan Education Department or the Shala Darpan portal.
Look for the “Latest News” or “Examination” section. Check school notice boards or ask your class teacher for updates.
Some district education office websites may also upload the key. Avoid unofficial websites as they may provide incorrect or fake keys.
Hrllo,
If you want to view the CM Shri School 2025-26 admission test result , visit the official website, click on the "CM Shri Schools Admission Test - 2025." then select the "CM Shri Admission Test Result 2025 - Merit List" link. Here you need to log in with your credentials and view or download the merit list pdf.
I hope it will clear your query!!
Hello aspirant,
Fairness and equal opportunity are guaranteed by the rigorous merit-based admissions process at CM Shri Schools. The merit list will be made public on edudel.nic.in, the official DOE website. Based on how well they did on the admission exam, students will be shortlisted.
To know the result, you can visit our site through following link:
https://school.careers360.com/articles/cm-shri-school-admission-test-result-2025
Thank you
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
As per latest syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters