Question

Write a proof for the sectional formula?

Answer 1

Step 1. Draw the figure on a coordinate plane that divides the line segment $AB$ internally in the ratio $m:n$

Consider the point $D\left(x,y\right)$ on a coordinate plane.

Step 2. Determine the coordinate $x$ :

From the figure, observe the following:

$$\begin{gathered} \angle DAC=\angle BDE\left[\mathrm{Corresponding}\mathrm{angles}\right] \\ \angle DCA=\angle BED={90}^{°} \end{gathered}$$

Since $∆ADC~∆BDC$, by AA similarity: $\frac{AB}{DB}=\frac{AC}{DE}=\frac{DC}{BC}=\frac{m}{n}\dots \left(1\right)$

The coordinates of $AC,DE,DC,BE$ are as follows:

$\begin{array}{rcl}AC& =& x-x_1\dots \left(2\right)\\ DE& =& x_2-x\dots \left(3\right)\\ DC& =& y-y_1\dots \left(4\right)\\ BE& =& y_2-y\dots \left(5\right)\end{array}$

From $\left(1\right),\left(2\right),\mathrm{and}\left(3\right)$:

$\begin{array}{rcl}\frac{x-x_1}{x_2-x}& =& \frac{m}{n}\\ & \Rightarrow & nx-n{x}_1=m{x}_2-mx\\ & \Rightarrow & nx+mx=m{x}_2+n{x}_1\\ & \Rightarrow & x=\frac{m{x}_2+n{x}_1}{m+n}\end{array}$

Step 3. Determine the coordinate $y$:

From $\left(1\right),\left(4\right),\mathrm{and}\left(5\right)$:

$\begin{array}{rcl}\frac{y-y_1}{y_2-y}& =& \frac{m}{n}\\ & \Rightarrow & ny-n{y}_1=m{y}_2-my\\ & \Rightarrow & ny+my=m{y}_2+n{y}_1\\ & \Rightarrow & y=\frac{m{y}_2+n{y}_1}{n+m}\end{array}$

Step 4: Determine the section formula :

$D\left(x,y\right)=\left(\begin{array}{rcl}& & \frac{m{x}_2+n{x}_1}{m+n}\end{array},\begin{array}{rcl}& & \frac{m{y}_2+n{y}_1}{n+m}\end{array}\right)$

Hence, the point $D\left(x,y\right)$ on a coordinate plane that divides the line segment $AB$ internally in the ratio $m:n$ is $D\left(x,y\right)=\left(\begin{array}{rcl}& & \frac{m{x}_2+n{x}_1}{m+n}\end{array},\begin{array}{rcl}& & \frac{m{y}_2+n{y}_1}{n+m}\end{array}\right)$.