Consider a line segment $\overline{AB}$ with end points A($x_1$, $y_1$ ) and B($x_2$, $y_2$). Let M(x, y) be a point on the line segment $\overline{AB}$ which divides AB in m:n ratio. Let C, D, and E be the feet of perpendiculars of A, B, and M, respectively. Join AC, ME, and BD. Let F be the foot of the perpendicular of A on ME, and G be the foot of the perpendicular of M on BD.

From the figure, it is clear that

AF = $x - x_1$, MF = $y - y_1$, MG = $x_2 - x$, and BG = $y_2 - y$

Moreover, $\Delta$AFM and $\Delta$MGB are similar triangles. Thus, we have

$\frac{AM}{MB}$ = $\frac{AF}{MG}$ = $\frac{MF}{BG}$$\rightarrow$

$\frac{m}{n}$ =

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

$\frac{y - y_1}{y_2 - y}$$\rightarrow$m($x_2 - x$) = n($x - x_1$ ) and m($y_2 - y$) = n($y - y_1$)

$\rightarrow$m$x_2 - mx$ = $nx-nx_1$ and m$y_2$ - my = $ny - ny_1$

$\rightarrow$(m+n)x = $mx_2$ + $nx_1$ and (m+n)y = $my_2$ + $ny_1$

$\rightarrow$x =

$\frac{mx_2 + nx_1}{m+n}$ and y =

$\frac{my_2 + ny_1}{m+n}$So, we have

M(x, y) = (

$\frac{mx_2 + nx_1}{m+n}$,

$\frac{my_2 + ny_1}{m+n}$)

If M is the mid-point of $\overline{AB}$, then m=n. Thus,

Mid-point of $\overline{AB}$ = ($\frac{x_1 + x_2}{2}$, $\frac{y_1 + y_2}{2}$) ### Solved Example

**Question: **Find the mid-point of the line segment $\overline{AB}$ with end point A(-3, 4) and B(7, 2).

** Solution: **

Given that

A($x_1$, $y_1$) = (-3, 4)

B($x_2$, $y_2$) = (7, 2)

Thus, the mid-point of $\overline{AB}$ is given by

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

=($\frac{-3+7}{2}$, $\frac{4+2}{2}$)

=($\frac{4}{2}$, $\frac{6}{2}$)

=(2, 3)