Section Formula 

The Section Formula in coordinate geometry is used to find the coordinates of a point that divides a line segment joining two given points in a specific ratio. It applies to both internal and external divisions. If a point divides the segment in the ratio m:n, the formula gives its exact location. This concept is fundamental in geometry, with applications in mathematics, physics, engineering, and computer graphics. It helps to determine positions and relationships between points in a coordinate plane.

1.0What is the Section Formula?

The Section Formula helps us determine the coordinates of a point that divides a line segment internally or externally in a given ratio.

Suppose you have two points:

• A(x_1, y_1)
• B(x_2, y_2)

And a point P(x, y) divides the line segment AB in the ratio m: n.

2.0Internal Division Formula

If point P divides the line internally, then the coordinates of P are given by:

$P(x,y)=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

This means that the x and y coordinates of the point are essentially weighted averages of the endpoints, depending on the ratio m:n.

3.0Midpoint Formula

A special case of the Section Formula is when the line segment is divided exactly in the middle—that is, in the ratio 1:1. This gives us the Midpoint Formula.

If $A(x_1,y_1)andB(x_2,y_2)$ are the endpoints of a line segment, then the midpoint M is given by:

This formula simply finds the average of the x-coordinates and y-coordinates, giving the exact center point of the segment AB.

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

4.0External Division Formula

In some problems, the point divides the line externally, i.e., it lies outside the segment AB. In such a case, the coordinates are:

$P(x,y)=\left(\frac{m{x}_2-n{x}_1}{m-n},\frac{m{y}_2-n{y}_1}{m-n}\right)$

Just be cautious—since this involves subtraction, make sure you're applying the ratio correctly, especially when m=nm = n, as this would lead to division by zero (undefined point).

5.0Solved Examples on Section Formula 

1. Internal Division

Example 1. Find the coordinates of the point that divides the line segment joining A(4, 2) and B(10, 6) internally in the ratio 2:3.

Solution:

Using Internal Divison Formula 

$P(x,y)=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

$x=\frac{(2\times 10)+(3\times 4)}{2+3}=\frac{20+12}{5}=\frac{32}{5}=6.4$

$y=\frac{(2\times 6)+(3\times 2)}{2+3}=\frac{12+6}{5}=\frac{18}{5}=3.6$

Point = (6.4, 3.6)

Example 2. A point P divides the line segment joining A(-1, 4) and B(5, -2) in the ratio 3:1 internally.

Solution:

Using Internal Divison Formula 

$P(x,y)=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

$x=\frac{(3\times 5)+(1\times -1)}{3+1}=\frac{15-1}{4}=\frac{14}{4}=3.5$

$y=\frac{(3\times -2)+(1\times 4)}{4}=\frac{-6+4}{4}=\frac{-2}{4}=-0.5$

Point = (3.5, -0.5)

Example 3. The point P(x, y) divides the line joining A(2, −3) and B(8, 5) in the ratio 1:2.

Solution:

Using Internal Divison Formula 

$P(x,y)=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

$x=\frac{(1\times 8)+(2\times 2)}{1+2}=\frac{8+4}{3}=\frac{12}{3}=4$

$y=\frac{(1\times 5)+(2\times -3)}{3}=\frac{5-6}{3}=\frac{-1}{3}$

Point = $(4,\frac{-1}{3})$

2. Midpoint

Example 4. Find the midpoint of the line segment joining A(7, −3) and B(−5, 9).

Solution:

Using Midpoint Formula

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

$x=\frac{7+(-5)}{2}=\frac{2}{2}=1$

$y=\frac{-3+9}{2}=\frac{6}{2}=3$

Midpoint = (1, 3)

Example 5. The midpoint is M(4, 5), one end is A(2, 3). Find the other end B.

Solution:

B = (x, y), then:

$\frac{2+x}{2}=4\Rightarrow x=6\frac{3+y}{2}=5\Rightarrow y=7$

Point B = (6, 7)

Example 6. Show that the midpoint formula is a special case of Section Formula with a ratio of 1:1.

Solution:
Using section formula:

$x=\frac{1\cdot x_2+1\cdot x_1}{1+1}=\frac{x_1+x_2}{2}y=\frac{y_1+y_2}{2}$

Which is exactly the midpoint formula. 

3. External Division

Example 7. Find the coordinates of the point that divides A(3,−1)A(3, -1) and B(9,5)B(9, 5) externally in the ratio 2:12:1.

Solution:

Using Extermal Divison Formula

$P(x,y)=\left(\frac{m{x}_2-n{x}_1}{m-n},\frac{m{y}_2-n{y}_1}{m-n}\right)$

$x=\frac{(2\cdot 9)-(1\cdot 3)}{2-1}=\frac{18-3}{1}=15$

$y=\frac{(2\cdot 5)-(1\cdot -1)}{1}=\frac{10+1}{1}=11$

Point = (15, 11)

Example 8. A point PP divides A(−6, 2) and B(4, −8) externally in the ratio 3:2.

Solution:

Using Extermal Divison Formula

$P(x,y)=\left(\frac{m{x}_2-n{x}_1}{m-n},\frac{m{y}_2-n{y}_1}{m-n}\right)$

$x=\frac{(3\cdot 4)-(2\cdot -6)}{3-2}=\frac{12+12}{1}=24$

$y=\frac{(3\cdot -8)-(2\cdot 2)}{1}=\frac{-24-4}{1}=-28$

Point = (24, -28)

Example 9. Find a point P on the line joining A(5, 2) and B(−1, −4) such that P divides the line externally in the ratio 5:3.

Solution:

Using Extermal Divison Formula

$P(x,y)=\left(\frac{m{x}_2-n{x}_1}{m-n},\frac{m{y}_2-n{y}_1}{m-n}\right)$

$x=\frac{(5\cdot -1)-(3\cdot 5)}{5-3}=\frac{-5-15}{2}=\frac{-20}{2}=-10$

$y=\frac{(5\cdot -4)-(3\cdot 2)}{2}=\frac{-20-6}{2}=\frac{-26}{2}=-13$

Point = (-10, -13)

6.0Why is the Section Formula Important?

The Section Formula is more than just a classroom concept:

• It helps in geographical positioning (like GPS triangulation).
• Used in computer graphics for positioning and animation.
• Plays a role in engineering and architecture for design modeling.
• Crucial for advanced math topics like vectors and calculus.