Section Formula – Formula, Derivation & Examples

In coordinate geometry, the section formula is used to find the ratio in which a line segment is divided by a point internally or externally. It is used to find out the centroid, incenter, and excenters of a triangle. In physics also, it is used to find the centre of mass of systems, equilibrium points, etc.

Let’s understand what is section formula, its derivation, and its uses with examples.

What is Section Formula?

The section formula is used to determine the coordinates of a point that divides a line segment into two segments. It is a handy tool used to find the coordinates of the point dividing the line segment in some ratio. The section formula can also be used to find the midpoint of a line segment and for the derivation of the midpoint formula as well.

The section formula states that if a point $\text{P}(x,y)$ divides the line segment with marked points as $\text{A} (x_1, y_1)$ and $\text{B}(x_2,y_2)$ in a ratio $m : n$,  then the coordinates of $\text{P}(x,y) = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)$.

Types of Section Formula

There are two forms of section formula.

• Internal Section Formula
• External Section Formula

Internal Section Formula

The internal section formula is used when the line segment is divided internally by a point in a particular ratio. 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.

The above figure shows a line segment $\text{AB}$ divided by a point $\text{P}(x, y)$ internally in a ratio such that $\text{AP}: \text{PB} = m:n$. Then the coordinates of the point $\text{P}$ in terms of the given ratio and for the given coordinates of the points $\text{A}(x_1, y_1)$, and $\text{B}(x_2, y_2)$ is expressed as $\text{P}(x,y) = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)$.

where

• $x$ and $y$ are the coordinates of point $\text{P}$
• $(x_1, y_1)$ are the coordinates of point $\text{A}$
• $(x_2, y_2)$ are the coordinates of the point $\text{B}$
• $m$ and $n$ are the ratio values in which $\text{P}$ divides the line internally

Derivation of Section Formula

To derive the section formula, we will the concept of two similar right-angled triangles. The hypotenuse of these two right angles is in the given ratio $m: n$. For this, we need to do some construction. Consider the point $\text{P}(x, y)$ on the coordinate plane, which divides the line segment $\text{AB}$ internally. Extend the horizontal lines and vertical lines from the three given points to form two right-angled triangles $\text{AQP}$ and $\text{PRB}$, as shown in the figure below.

In the two right triangles $\text{AQP}$ and $\text{PRB}$,

$\angle \text{PAQ} = \angle \text{BPR}$ (corresponding angles)

$\angle \text{PQA} = \angle \text{BRP} = 90^{\circ}$

Therefore the triangles $\text{AQP}$ and $\text{PRB}$ are similar by $\text{AA}$ similarity rule.

Thus

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

Now using coordinates, we have

$\text{AQ} = x – x_1$ ———————————(2)

$\text{PR} = x_2 – x$ ______________________________(3)

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

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

$=> n(x – x_1) = m(x_2 – x)$

$=> nx – nx_1 = mx_2 – mx$

$=> nx + mx = mx_2 + nx_1$

$=> (m + n)x = mx_2 + nx_1$

$=> x = \frac{mx_2 + nx_1}{m + n}$

Similarly we can derive $y = \frac{my_2 + nx_1}{m + n}$

Mid-Point Formula

From the above section formula, we can derive the midpoint formula. A point is called a midpoint of the join of two points, if it divides the line segment in a ratio $1 : 1$.

Replacing $m$ and $n$ in the section formula with $1$ and $1$, we get $\text{P}(x,y) = \left(\frac{1 \times x_2 + 1 \times x_1}{1 + 1}, \frac{1 \times y_2 + 1 \times y_1}{1 + 1} \right) $

$=> \text{P}(x,y) = \left(\frac{x_2 + x_1}{2}, \frac{y_2 + y_1}{2} \right)$

$=> \text{P}(x,y) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$ is the midpoint formula used to find the midpoint of a line segment joining the points $\text{A}(x_1, y_1)$, and $\text{B}(x_2, y_2)$.

Examples of Internal Section Formula

Example 1: Find the coordinates of the point which divides the join of $(–1, 7)$ and $(4, –3)$ in the ratio $2 : 3$.

According to section formula a line joining the points $\text{A}(x_1, y_1)$ and $\text{B}(x_2, y_2)$ is divided in a ratio $m : n$ by a point $\text{P}\left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)$.

Here $(x_1, y_1) = (–1, 7)$, $(x_2, y_2) = (4, -3)$, and $m : n = 2 : 3$.

Therefore $x = \frac{m x_2 + n x_1}{m + n} = \frac{2 \times 4 + 3 \times (-1)}{2+ 3}$

$=>x = \frac{8 – 3}{5}$

$=>x = \frac{5}{5} = 1$

$y = \frac{m y_2 + n y_1}{m + n} = \frac{2 \times (-3) + 3 \times 7}{2 + 3}$

$=> y = \frac{-6 + 21}{2 + 3}$

$=> y = \frac{15}{5} = 3$

Thus coordinates of a point dividing a line segment joining the points $(–1, 7)$ and $(4, –3)$ in the ratio $2 : 3$ is $(1, 3)$.

Example 2: Find the coordinates of the points of trisection of the line segment joining $(4, –1)$ and $(–2, –3)$.

Let $\text{P}(a, b)$ and $\text{Q}(c, d)$ be the two points lying on the join of the points $\text{A}(4, -1)$ and $\text{B}(-2, -3)$ such that $\text{AP} : \text{PQ} : \text{QB} = 1 : 1 : 1$.

Therefore the point $\text{P}(a, b)$ divides the join of the points $\text{A}(4, -1)$ and $\text{B}(-2, -3)$ in ratio $1 : 2$ and the point $\text{Q}(c, d)$ divides the join of the points $\text{A}(4, -1)$ and $\text{B}(-2, -3)$ in ratio $2 : 1$.

$a = \left(\frac{1 \times (-2) + 2\times 4}{1 + 2} \right)$

$=> a = \left(\frac{-2 + 8}{3} \right)$

$=> a = \left(\frac{6}{3} \right) = 2$

$b = \left(\frac{1 \times (-3) + 2\times (-1)}{1 + 2} \right)$

$=>b = \left(\frac{-3 -2}{3} \right) = -\frac{5}{3}$

$c = \left(\frac{2 \times (-2) + 1\times 4}{2 + 1} \right)$

$=> c = \left(\frac{-4 + 4}{3} \right) = 0$

$d = \left(\frac{2 \times (-3) + 1\times (-1)}{2 + 1} \right)$

$=>d = \left(\frac{-6 -1}{3} \right) = – \frac{7}{3}$

Thus the points of trisection of the line segment joining $(4, –1)$ and $(–2, –3)$ are $\left(2, -\frac{5}{3} \right)$ and $\left(0, – \frac{7}{3} \right)$.

Example 3: Find the ratio in which the line segment joining the points $(– 3, 10)$ and $(6, – 8)$ is divided by $(– 1, 6)$.

Here the two endpoints of the line are $\text{A}(– 3, 10)$ and $\text{B}(6, – 8)$ and the point lying on the line joining these two points is $\text{P}(-1, 6)$.

Let the ratio be $m : n$

Therefore $-1 = \frac{m \times 6 + n \times (-3)}{m + n}$

$=> -1 = \frac{6m – 3n}{m + n}$

$=> -1(m + n) = 6m – 3n$

$=> -m – n = 6m – 3n$

$=> – n + 3n = 6m + m$

$=> 2n = 7m$

$=> \frac{m}{n} = \frac{2}{7}$

Thus the ratio in which the line segment joining the points $(– 3, 10)$ and $(6, – 8)$ is divided by $(– 1, 6)$ is $2 : 7$

Example 4: If $(1, 2)$, $(4, y)$, $(x, 6)$ and $(3, 5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$

Let the vertices of the parallelogram are $\text{A}(1, 2)$, $\text{B}(4, y)$, $\text{C}(x, 6)$ and $\text{D}(3, 5)$.

Therefore the diagonals of the parallelogram are $\text{AC}$ and $\text{BD}$.

Since the diagonals of a parallelogram bisect each other, it means the midpoints of $\text{AC}$ and $\text{BD}$ coincide, i.e., the midpoint of $\text{AC}$ and $\text{BD}$ are same.

Midpoint of $\text{AC}$ = $\left(\frac{1 + x}{2}, \frac{2 + 6}{2} \right)$

Midpoint of $\text{BD}$ = $\left(\frac{4 + 3}{2}, \frac{y + 5}{2} \right)$

Therefore $\left(\frac{1 + x}{2}, \frac{2 + 6}{2} \right) = \left(\frac{4 + 3}{2}, \frac{y + 5}{2} \right)$

$=> \left(\frac{1 + x}{2}, \frac{8}{2} \right) = \left(\frac{7}{2}, \frac{y + 5}{2} \right)$

$=> \left(\frac{1 + x}{2}, 4 \right) = \left(\frac{7}{2}, \frac{y + 5}{2} \right)$

$=> \frac{1 + x}{2} = \frac{7}{2}$ and $\frac{y + 5}{2} = 4$

$=> 1 + x = 7$ and $y + 5 = 4 \times 2$

$=> x = 7 – 1$ and $y + 5 = 8$

$=> x = 6$ and $y = 8 – 5$

$=> x = 6$ and $y = 3$

Example 5: Line $2x+y−4=0$ divides the line segment joining the points $\text{A}(2,−2)$ and $\text{B}(3,7)$. Find the ratio of line segment in which the line is dividing?

Let a point $\text{P}(a, b)$ divides the line segment joining the points $\text{A}(2,−2)$ and $\text{B}(3,7)$ in ratio $k : 1$. Therefore $a = \frac{k \times 3 + 1 \times 2}{k + 1}$ and $b = \frac{k \times 7 + 1 \times (-2)}{k + 1}$

$=> a = \frac{3k + 2}{k + 1}$ and $b = \frac{7k – 2}{k + 1}$

Since the point $(a, b)$ lies on the line $2x+y−4=0$, therefore $2 \times \frac{3k + 2}{k + 1} + \frac{7k – 2}{k + 1} − 4=0$

$=> 2 \times \frac{3k + 2}{k + 1} + \frac{7k – 2}{k + 1} = 4$

$=> 2(3k + 2) + (7k – 2) = 4(k + 1)$

$=> 6k + 4 + 7k – 2 = 4k + 4$

$=> 13k + 2 = 4k + 4$

$=> 13k – 4k = 4 – 2$

$=> 9k = 2 => k = \frac{2}{9}$

External Section Formula

The external section formula is used when the line segment is divided externally by the point in the given ratio. 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.

The above figure shows a line segment $\text{AB}$ divided by a point $\text{P}(x, y)$ externally in a ratio such that $\text{AP}: \text{PB} = m:n$.

Then the coordinates of the point $\text{P}$ in terms of the given ratio and for the given coordinates of the points $\text{A}(x_1, y_1)$, and $\text{B}(x_2, y2)$ is expressed as $\text{P}(x,y) = \left(\frac{m x_2 – n x_1}{m – n}, \frac{m y_2 – n y_1}{m – n} \right)$.

where

• $x$ and $y$ are the coordinates of point $\text{P}$
• $(x_1, y_1)$ are the coordinates of point $\text{A}$
• $(x_2, y_2)$ are the coordinates of the point $\text{B}$
• $m$ and $n$ are the ratio values in which $\text{P}$ divides the line internally

Examples of External Section Formula

Example 1: $\text{A} (4, 5)$ and $\text{B} (7, -1)$ are two given points, and a point $\text{C}$ divides the line-segment $\text{AB}$ externally in the ratio $4 : 3$. Find the coordinates of $\text{C}$.

According to external section formula a line joining the points $\text{A}(x_1, y_1)$ and $\text{B}(x_2, y_2)$ is divided in a ratio $m : n$ by a point $\text{P}\left(\frac{m x_2 – n x_1}{m – n}, \frac{m y_2 – n y_1}{m – n} \right)$.

Here $(x_1, y_1) = (4, 5)$, $(x_2, y_2) = (7, -1)$, and $m : n = 4 : 3$.

Therefore $x = \frac{4 \times 7 – 3 \times 4}{4 – 3}$

$=>x = \frac{28 – 12}{1} = 16$

And $y = \frac{4 \times (-1) – 3 \times 5}{4 – 3}$

$=> y = \frac{-4 – 15}{1} = -19$

Thus $(16, -19)$ divides the line segment joining the points $\text{A} (4, 5)$ and $\text{B} (7, -1)$ in the ratio $4 : 3$.

Practice Problems

1. Find the coordinates of the point which divides the line segment joining the points $(4,6)$ and $(-5,-4)$ internally in the ratio $3:2$.
2. Find the midpoint of the line segment $\text{AB}$ which joins $\text{A} (4,8)$ and $\text{B}(2,4)$.
3. $\text{A}(2, 7)$ and $\text{B}(–4, –8)$ are coordinates of the line segment $\text{AB}$. There are two points that trisected the segment. Find the coordinates of them.
4. The $4$ vertices of a parallelogram are $\text{A}(-2, 3)$, $\text{B}(3, -1)$, $\text{C}(p, q)$ and $\text{D}(-1, 9)$. Find the value of  $p$ and $q$.
5. Find the coordinates of a point $\text{M}$ which divides a line segment $\text{PQ}$ in the ratio $2:3$. $\text{PQ}$ line segment joins the points $\text{P} (2,1)$ and $\text{Q} (-3,6)$. Does the point $\text{M}$ lie on the line $5y – x = 15$?

FAQs

Conclusion

Section formula is one of the most important formulas of coordinate geometry. It is used to find out the centroid, incenter, and excenters of a triangle. In physics also, it is used to find the centre of mass of systems, equilibrium points, etc. The section formula is given by $\left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)$, where a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is divided by a point in a ratio $m : n$.

Recommended Reading

• Coordinate Geometry – Meaning, Concept & Overview
• Distance Formula in Coordinate Geometry