Section Formula - Definition, Formulas, Proof and Examples

Section Formula in Coordinate Geometry – Definition, Formula and Applications

The Section Formula is a fundamental concept in coordinate geometry and vector algebra. It is used to find the exact coordinates of a point that divides a line segment in a given ratio. This formula not only helps in calculating the ratio in which a point divides a line but also helps in finding the coordinates of that point accurately. One of its most common applications is finding the midpoint of a line segment, where the ratio is 1:1. The Section Formula plays a vital role in geometry, physics, and engineering problems involving straight-line division and positioning of points.

Definition of Section Formula

When a point lies on a line segment and divides it into two parts in a given ratio, the formula used to find the coordinates of that point is called the Section Formula.

Let a point $P(x, y)$ divide the line segment joining
$A(x_1, y_1)$ and $B(x_2, y_2)$
in the ratio $m:n$. Then the coordinates of point $P$ are given by:

$x=\frac{m x_2+n x_1}{m+n}, \quad y=\frac{m y_2+n y_1}{m+n}$

This formula is mainly used to find the position of a point that divides a line internally or externally in a given ratio.

Types of Section Formula in Coordinate Geometry

The Section Formula is broadly divided into two important types:

1) Internal Section Formula

When a point divides the line segment internally between two given points in the ratio $m:n$, we use the internal section formula. It means the point lies between points $A$ and $B$. This is the most commonly used form of the section formula in mathematics.

2) External Section Formula

When a point divides the line segment externally in the ratio $m:n$, the point lies outside the line segment joining the two given points. This form is useful when the dividing point is beyond either endpoint of the line.

Let us describe each in detail:

Internal Division in Section Formula

When a point divides a line segment internally, we use the Internal Section Formula. This formula helps us find the coordinates of a point that lies between two given points on a line and divides the line segment in a given ratio. In Vector Algebra, this idea is used to express a point as a weighted combination of two position vectors.

Consider two points
$A(x_1, y_1)$ and $B(x_2, y_2)$ in a plane.
If a point $P(x, y)$ lies on the line segment $AB$ and is located somewhere between $A$ and $B$, then $P$ is said to divide $AB$ internally.

If $AP:PB = m:n$,
then point $P$ divides $AB$ internally in the ratio $m:n$.

The coordinates of point $P(x, y)$ are given by the Internal Section Formula:

$x=\frac{m x_2+n x_1}{m+n}, \quad y=\frac{m y_2+n y_1}{m+n}$

This formula is one of the most important results in Vector Algebra and Coordinate Geometry, especially for finding midpoints, trisection points, and proportional division of vectors.

External Division in Section Formula

When a point divides a line segment externally, we use the External Section Formula. In this case, the point does not lie between the two given points but lies on the extension of the line beyond either end.

If point $P$ lies on the line joining $A$ and $B$ but not between them, and
$AP:BP = m:n$,
then $P$ is said to divide $AB$ externally in the ratio $m:n$.

The coordinates of point $P(x, y)$ are given by:

$x=\frac{m x_2-n x_1}{m-n}, \quad y=\frac{m y_2-n y_1}{m-n}$

This form is very useful in vector problems where the dividing point lies outside the given segment.

Section Formula for Midpoint

The Midpoint Formula is a special case of the internal division formula where the ratio is $1:1$. A midpoint is the point that lies exactly halfway between two points.

If $P$ is the midpoint of the line segment joining
$A(x_1, y_1)$ and $B(x_2, y_2)$, then:

$x=\frac{x_1+x_2}{2}, \quad y=\frac{y_1+y_2}{2}$

This is one of the most commonly used formulas in Vector Algebra and Coordinate Geometry.

Derivation of Section Formula

The Section Formula can be proved using the concept of similar triangles.

Let point $P(x, y)$ divide the line segment $AB$ internally in the ratio $m:n$.
Draw perpendiculars from points $A$, $P$, and $B$ to form two right-angled triangles $AQP$ and $PRB$.

In triangles $AQP$ and $PRB$:

∠PAQ = ∠BPR (corresponding angles)
∠PQA = ∠BRP = $90^\circ$

Hence, the triangles are similar by AA similarity.

So,

$\frac{AP}{PB}=\frac{AQ}{PR}=\frac{PQ}{BR}=\frac{m}{n}$ ...(1)

Using coordinates:

$AQ = x - x_1$ ...(2)
$PR = x_2 - x$ ...(3)

From (1), (2), and (3):

$\frac{x - x_1}{x_2 - x}=\frac{m}{n}$

Solving for $x$:

$x=\frac{m x_2+n x_1}{m+n}$ ...(A)

Similarly, for the $y$-coordinate:

$PQ = y - y_1$ ...(4)
$BR = y_2 - y$ ...(5)

$\frac{y - y_1}{y_2 - y}=\frac{m}{n}$

Solving for $y$:

$y=\frac{m y_2+n y_1}{m+n}$ ...(B)

From (A) and (B), the coordinates of point $P$ are:

$P(x, y)=\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)$

This proves the Internal Section Formula.

Important Results and Shortcuts in Section Formula

1. Sometimes, instead of using the ratio $m:n$, we use $\lambda:1$ for simplicity.
   The internal division formula becomes:
   $\left(\frac{\lambda x_2+x_1}{\lambda+1}, \frac{\lambda y_2+y_1}{\lambda+1}\right)$

2. If $\lambda > 0$, the division is internal.
   If $\lambda < 0$, the division is external.

3. The coordinates of the points that trisect the line segment $AB$ are:

   $\left(\frac{x_1+2x_2}{3}, \frac{y_1+2y_2}{3}\right)$
   and
   $\left(\frac{2x_1+x_2}{3}, \frac{2y_1+y_2}{3}\right)$

Solved Examples Based on Section Formula

Example 1: Let a triangle be bounded by the lines $\mathrm{L}_1: 2 x+5 y=10 ; \mathrm{L}_2:-4 x+3 y=12$ and the line $L_3$, which passes through the point $\mathrm{P}(2,3)$ intersects $\mathrm{L}_2$ at $A$ and $L_1$ at $B$ . If the point P divides the line-segment AB , internally in the ratio $1: 3$,then the area of the triangle is equal to [JEE MAINS 2022]

Solution

Let A be $\left(\mathrm{a}, 4+\frac{4 \mathrm{a}}{3}\right) \quad\left(\because \mathrm{A}\right.$ lies on $\left.\mathrm{L}_2\right)$

and B be $\left(\mathrm{b}, 2-\frac{2}{5} \mathrm{~b}\right) \quad\left(\because \mathrm{B}\right.$ lies on $\left.\mathrm{L}_1\right)$

Using the section formula for point P

$2=\frac{3 \mathrm{a}+\mathrm{b}}{4}$ and $3=\frac{3\left(4+\frac{4 \mathrm{a}}{3}\right)+\left(2-\frac{2 \mathrm{~b}}{5}\right)}{4}$

$\Rightarrow 3 \mathrm{a}+\mathrm{b}=8$ and $4 \mathrm{a}-\frac{2 \mathrm{~b}}{5}=-2$

$\Rightarrow \quad \mathrm{a}=\frac{3}{13}, \mathrm{~b}=\frac{95}{13}$

$\therefore \quad \mathrm{A}$ is $\left(\frac{3}{13}, \frac{56}{13}\right), \mathrm{B}$ is $\left(\frac{95}{13}, \frac{-12}{13}\right)$

$\therefore$ Area of $\triangle \mathrm{ABC}=\frac{1}{2}\left|\begin{array}{ccc}\frac{3}{13} & \frac{56}{13} & 1 \\ \frac{95}{13} & \frac{-12}{13} & 1 \\ \frac{-15}{13} & \frac{32}{13} & 1\end{array}\right|$

$=\frac{132}{13}$

Hence, the correct answer is $\frac{132}{13}$.

Example 2: The locus of a point that divides the line segment joining the point $(0,-1)$ and a point on the parabola,$x^2=4 y$ internally in the ratio $1: 2$ is : [JEE MAINS 2020]

Solution : Let point $P$ be $\left(2 t, t^2\right)$ and $Q$ be $(h, k)$

$
h=\frac{2 t}{3}, k=\frac{-2+t^2}{3}
$

Hence locus is $3 k+2=\left(\frac{3 h}{2}\right)^2 \Rightarrow 9 x^2=12 y+8$

Hence, the answer is 9 x^2-12 y=8

Example 3: Two sides of a parallelogram are along the lines, $x+y=3$ and $x-y+3=0$ . If its diagonals intersect at $(2,4)$ , then one of its vertex is: [JEE MAINS 2019]

Solution: Mid-point formula is given by

$\begin{aligned} & x=\frac{x_1+x_2}{2} \\ & y=\frac{y_1+y_2}{2}\end{aligned}$

If the point P(x,y) is the midpoint of the line joining $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$.

Two lines

$x+y=3$ and $x-y=-3$ intersects at A (0,3)

Point C is $\left(x_1, y_1\right)$

So,

$\frac{x_1+0}{2}=2 \quad \frac{y_1+4}{2}=4$

$=>C\left(x_1, y_1\right)=C(4,5)$

So, the Equation of BC is $x-y=-1$

and the equation of CD is $x+y=9$

Solve $x+y=9$ and $x-y=-3$

$D(3,6)$

Hence, the answer is $(3,6)$.

Example 4: If in a parallelogram ABDC, the coordinates of A, B, and C are respectively $(1,2),(3,4)$ and $(2,5)$. then the equation of the diagonal AD is: [JEE MAINS 2019]

Solution: Mid-point formula is given by

$\begin{aligned} & x=\frac{x_1+x_2}{2} \\ & y=\frac{y_1+y_2}{2}\end{aligned}$

If the point $P(x, y)$ is the midpoint of the line joining $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $B\left(x_2, y_2\right)$.

Two–point form of a straight line -

$y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)$

The lines pass through $\left(x_1 y_1\right)$ and $\left(x_2 y_2\right)$.

As BD and AC are parallel

$\frac{n-4}{m-3}=\frac{5-2}{2-1}$

$n-4=3(m-3)$..............................(1)

As AB and CD are parallel

$\frac{n-5}{m-2}=\frac{4-2}{3-1}=\frac{2}{2}=1$

$n-5=(m-2)$..............................(2)

Solving (1) and (2)

m=4 and n=7

$\overrightarrow{D A}$ is $(y-2)=\left(\frac{7-2}{4-1}\right)(x-1)$

$=>5 x-3 y+1=0$

Hence, the answer is $5 x-3 y+1=0$.

Example 5: If a circle C, whose radius is 3, touches externally the circle, $x^2+y^2+2 x-4 y-4=0$ at the point $(2,2)$, then the length of the intercept cut by this circle C, on the x-axis, is equal to [JEE MAINS 2018]

Solution: The general form of a circle - $x^2+y^2+2 g x+2 f y+c=0$

centre = $(-g,-f)$

radius = $\sqrt{g^2+f^2-c}$

$S=x^2+y^2+2 g x+2 f y+c=0$

equation of

$C_1 i s(x-5)^2+(y-2)^2=3^2$

$x^2+y^2-10 x-4 y+20=0$

X- intercept = $\left.2 \sqrt{(} g^2-c\right)=2 \sqrt{(25-20)}$

$=2 \sqrt{5}$

Hence, the answer is $2 \sqrt{5}$.

List of Topics Related to Section Formula

This section gives a quick overview of all the important topics covered in Straight Lines: section formula, so you can clearly see what concepts you need to study and how they are connected with each other.

Equation of Straight Line

Pair of Straight Lines

Family of Lines

NCERT Resources

This section includes useful NCERT-based materials such as notes, solutions, exemplar solutions that are directly aligned with the syllabus and help in building strong fundamentals.

NCERT Maths Class 11th Notes for Chapter 10 - Straight Lines

NCERT Maths Class 11th Solutions for Chapter 10 - Straight Lines

NCERT Maths Class 11th Exemplar Solutions for Chapter 10 - Straight Lines

Practice Questions based on the Section Formula

This section contains carefully selected questions to help you apply concepts, improve problem-solving skills, and gain confidence through regular practice.

Section Formula- Practice Question MCQ

We have provided below the practice questions based on the topics related to the section formula: