ICSE Class 10 Maths Section Formula Solutions Guide

Written by Greya LakshmiPublished on 19 May 2026Updated on 9 June 2026

ICSE Class 10 Maths Chapter 13 Section Formula

ICSE Class 10 Maths Chapter 13, Section Formula and Mid-Point Formula, teaches how to find the coordinates of a point that divides a line segment in a given ratio. For Concise Mathematics Selina Solutions Class 10 ICSE Chapter 13 Section Formula and Mid-Point Formula, the safest method is: identify $A(x_1,y_1)$, $B(x_2,y_2)$, write $AP:PB=m:n$ , substitute carefully, and simplify the ordered pair.

Concept snapshot

Section formula is a balance rule. If $P$ divides $AB$ in the ratio $AP:PB=m:n$ , the first ratio number $m$ pulls the answer towards $B(x_2,y_2)$, while $n$ pulls it towards $A(x_1,y_1)$. This explains why the formula uses $mx_2+nx_1$ , not $mx_1+nx_2$ .

Formula reference for Chapter 13

Case	Formula	Use
Internal section formula	$P\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)$	When $AP:PB=m:n$ .
Mid-point formula	$M\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$	When the ratio is $1:1$ .
Centroid	$G\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)$	For the centroid of a triangle.
Point on $x$ -axis	$(x,0)$	Use $y=0$ .
Point on $y$ -axis	$(0,y)$	Use $x=0$ .

Worked examples for section formula, mid-point and centroid

Worked Example 1: Divide $A(2,-1)$ and $B(8,5)$ in the ratio $1:2$

Step 1: Here $m=1$ , $n=2$ , $A(2,-1)$, and $B(8,5)$.

Step 2: Apply the internal section formula.

$P=\left(\dfrac{1\cdot8+2\cdot2}{1+2},\dfrac{1\cdot5+2\cdot(-1)}{1+2}\right)$

$P=\left(\dfrac{12}{3},\dfrac{3}{3}\right)=(4,1)$

Final answer: $P=(4,1)$.

Worked Example 2: Find the mid-point of $(-4,7)$ and $(6,-3)$

Step 1: Use the mid-point formula.

$M=\left(\dfrac{-4+6}{2},\dfrac{7+(-3)}{2}\right)$

$M=\left(\dfrac{2}{2},\dfrac{4}{2}\right)=(1,2)$

Final answer: $M=(1,2)$.

Worked Example 3: Find the centroid of $(1,2)$, $(7,2)$, and $(4,8)$

Step 1: Use the centroid formula.

$G=\left(\dfrac{1+7+4}{3},\dfrac{2+2+8}{3}\right)$

$G=(4,4)$

Final answer: The centroid is $(4,4)$.

Exercise 13(A) model solutions

Question 1(a): $P$ divides $A(1,3)$ and $B(5,9)$ in the ratio $1:2$

Step 1: Take $m=1$ , $n=2$ .

$P=\left(\dfrac{1\cdot5+2\cdot1}{3},\dfrac{1\cdot9+2\cdot3}{3}\right)$

$P=\left(\dfrac{7}{3},5\right)$

Final answer: $P=\left(\dfrac{7}{3},5\right)$.

Question 1(b): $A(2,4)$, $B(6,12)$, and $P=(3,x)$

Step 1: Let $AP:PB=m:n$ .

$3=\dfrac{6m+2n}{m+n}$

$3m+3n=6m+2n\Rightarrow n=3m$

Final answer: $AP:PB=1:3$ . Also, $x=6$ .

Question 4: The $x$ -axis divides the join of $(4,3)$ and $(2,-6)$

Step 1: A point on the $x$ -axis is $(x,0)$.

$0=\dfrac{-6m+3n}{m+n}\Rightarrow 6m=3n\Rightarrow m:n=1:2$

$x=\dfrac{1\cdot2+2\cdot4}{3}=\dfrac{10}{3}$

Final answer: Ratio $1:2$ , point $\left(\dfrac{10}{3},0\right)$.

Question 8: $P$ lies on $A(4,3)$, $B(-2,6)$, and $5AP=2BP$

Step 1: Convert $5AP=2BP$ into $\dfrac{AP}{BP}=\dfrac{2}{5}$ , so $AP:PB=2:5$ .

$P=\left(\dfrac{2\cdot(-2)+5\cdot4}{7},\dfrac{2\cdot6+5\cdot3}{7}\right)$

$P=\left(\dfrac{16}{7},\dfrac{27}{7}\right)$

Final answer: $P=\left(\dfrac{16}{7},\dfrac{27}{7}\right)$.

Quick answer index

Question	Answer
1(a)	$\left(\dfrac{7}{3},5\right)$
1(b)	$AP:PB=1:3$
1(c)	$5:3$ , point $(7,9)$
1(d)	$3:5$
1(e)	$2:5$ , point $(0,3)$
2	$2:3$ , $a=2$
3	$3:2$ , $a=-\dfrac{2}{5}$
4	$1:2$ , point $\left(\dfrac{10}{3},0\right)$
5	$4:3$ , point $(0,3)$
6	$B=(3,-6)$, $D=(1,-2)$
7	$\left(-\dfrac{11}{5},\dfrac{14}{5}\right)$
8	$\left(\dfrac{16}{7},\dfrac{27}{7}\right)$
9	$5:3$ , point $(2,4)$
10	$3:5$ , point $\left(\dfrac{21}{4},2\right)$

Examiner’s mindset

For coordinate geometry, a board-style answer should show the formula, correct substitution, algebra, and final ordered pair. If the question asks for both ratio and point, give both. Do not write only the option letter while practising MCQs.

Common mistakes students make

Reversing the ratio: For $AP:PB=m:n$ , use $mx_2+nx_1$ , not $mx_1+nx_2$ .
Swapping axes: On the $x$ -axis, $y=0$ ; on the $y$ -axis, $x=0$ .
Using $5:2$ for $5AP=2BP$ : The correct ratio is $AP:PB=2:5$ .
Losing signs: Keep negative coordinates such as $-1$ and $-6$ throughout the substitution.