Section Formula - Definition, Formulas, Proof and Examples

Section formula is used to find the ratio in which the line segment is divided by the point that lies externally or internally on a line. With the help of the section formula, we can also find the location of a point from where the line is divided. We can use the Section formula to find the centroid, incentre, and excentre of the triangle.

This Story also Contains

1. Section Formula
2. Definition of Section Formula
3. Internal division
4. External Division
5. Section Formula for a midpoint
6. Section Formula Derivation
7. Solved Examples Based on Section Formula

In this article, we will cover the concept of Section Formula. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of eleven questions have been asked on JEE MAINS ( 2013 to 2023) from this topic including one in 2014, one in 2015, one in 2016, three in 2018, three in 2019, one in 2020, and one in 2022.

Section Formula

The formula used to calculate the ratio in which a point divides the line segment is called the section formula. It is also used to calculate the coordinate of the point which divides the line into two parts. With the help of the section formula, we can calculate the midpoint of the line segment.

Definition of Section Formula

A point lying on the line divides the line segments into two parts, the formula used to calculate the coordinates of points is called the distance formula. If point P(x,y) divides the line segment joining A$
\left(x_1, y_1\right)
$ and B$
\left(x_2, y_2\right)
$ in the ration m:n. To find the coordinates, we use the section formula, which is mathematically expressed as:

$\mathbf{x}=\frac{\mathbf{m x}_2+\mathbf{n x}_{\mathbf{1}}}{\mathbf{m}+\mathbf{n}}, \quad \mathbf{y}=\frac{\mathbf{m y}_{\mathbf{2}}+\mathbf{n y}_{\mathbf{1}}}{\mathbf{m}+\mathbf{n}}$

Section formula in coordinate geometry is mainly divided into sub-formulas, which are:

1) Internal section formula

2) External section formula

Internal division

If the point divides the line segment internally we use the internal section formula. This formula is used to find the coordinates of a point, which falls between the two points and on the line joining these two points, in the given ratio.

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

Now, if AP:BP=m:n, then point P divides AB internally in the ratio m: n.

The coordinates of the point P (x, y) dividing the line segment joining the two points A (x₁, y₁) and B (x₂, y₂) internally in the ratio m: n is given by

$\mathbf{x}=\frac{\mathbf{m x}_2+\mathbf{n x}_{\mathbf{1}}}{\mathbf{m}+\mathbf{n}}, \quad \mathbf{y}=\frac{\mathbf{m y}_{\mathbf{2}}+\mathbf{n y}_{\mathbf{1}}}{\mathbf{m}+\mathbf{n}}$

External Division

If the point divides the line segment externally we use the external section formula. This formula is used to find the coordinates of the point on the line segment joining the two points and falling beyond the two points, in the given ratio.

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

The coordinates of the point P (x, y) dividing the line segment joining the two points A (x₁, y₁) and B (x₂, y₂) externally in the ratio m: n is given by

$\mathbf{x}=\frac{\mathbf{m x}_{\mathbf{2}}-\mathbf{n x}_{\mathbf{1}}}{\mathbf{m}-\mathbf{n}}, \quad \mathbf{y}=\frac{\mathbf{m y}_{\mathbf{2}}-\mathbf{n y}_{\mathbf{1}}}{\mathbf{m}-\mathbf{n}}$

Section Formula for a midpoint

The midpoint formula is used to find the coordinates of the midpoint of a line segment. Here, the ratio between the two parts is 1:1. A midpoint refers to a point that is exactly in the middle of the line segment.

If P is the midpoint of the line segment AB, then the ratio becomes equal, i.e. m = n, in this case, the coordinates of point P is

$\mathrm{x}=\frac{\mathrm{x}_1+\mathrm{x}_2}{2}, \mathrm{y}=\frac{\mathrm{y}_1+\mathrm{y}_2}{2}$

Section Formula Derivation

This formula can be proved by using two similar right-angled triangles. The hypotenuse of these two right angles is in the given ratio m: n. Let us prove the section formula using simple construction. Consider the point P (x, y) on the coordinate plane, which divides the line segment AB internally. Extend the horizontal line and vertical line from the three given points to form two right-angled triangles AQP and PRB, as shown in the following figure.

In the two right triangles AQP and PRB,

∠PAQ = ∠BPR (corresponding angles)

∠PQA = ∠ BRP = 90°

we can see that both triangles are similar by AA similarity.

So we can write:

$\mathrm{AP} / \mathrm{PB}=\mathrm{AQ} / \mathrm{PR}=\mathrm{PQ} / \mathrm{BR}=$ $\mathrm{m} / \mathrm{n} \ldots$ (1)

Now using coordinates,

AQ =$x-x_1 \ldots(2)$

PR = $x_2-x \ldots(3)$

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

$\left[\mathrm{x}-\mathrm{x}_1\right] /\left[\mathrm{x}_2-\mathrm{x}\right]=\mathrm{m} / \mathrm{n}$

On solving for x, we get

$\mathrm{x}=\left(\mathrm{mx}_2+\mathrm{nx_{1 }}\right) /(\mathrm{m}+\mathrm{n}) . .(\mathrm{A})$

Similarly for y, we have,

$P Q=$ $y-y_1 \ldots(4)$

$B R=$ $y_2-y \ldots(5)$

From equation (1), (4), and (5),

$y=\left(m y_2+n y_1\right) /(m+n) .$.

Hence, from equation (A), and (B), we get,

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

Important points:

1) If the ratio in which a given line segment is divided, is to be determined, then sometimes, for convenience (instead of taking the ratio m: n) we take the ratio λ: 1 and apply the formula for internal division $\left(\frac{\lambda x_2+x_1}{\lambda+1}, \frac{\lambda y_2+y_1}{\lambda+1}\right)$

2) If the value of $\lambda>0$, it is an internal division, otherwise, it is an external division (i.e. when $\lambda<0$ )

3) The coordinates of the point which trisects AB are $\left(\frac{x_1+2 x_2}{3}\right),\left(\frac{y_1+2 y_2}{3}\right)$ or $\left(\frac{2 x_1+x_2}{3}\right),\left(\frac{2 y_1+y_2}{3}\right)$

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 $2 \sqrt{5}$.