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 1:$\lambda$ ratio where $\lambda$ being a non-zero real number. Let C, D, and E be the foot 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 = $x_2 - x$,

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$

$\frac{1}{\lambda}$ =

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

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

$\rightarrow$ $x_2 - x$ = $\lambda x - \lambda x_1$ and $y_2 - y$ = $\lambda y - \lambda y_1$

$\rightarrow$ (1+ $\lambda$) x = $\lambda x_1$ + $x_2$ and (1 + $\lambda$) y = $\lambda y_1$ + $y_2$

$\rightarrow$ x =

$\frac{\lambda x_1 + x_2}{1 + \lambda}$ and y =

$\frac{\lambda y_1 + y_2}{1+ \lambda}$Thus, we have

M = (x, y) = (

$\frac{\lambda x_1 + x_2}{m + n}$,

$\frac{\lambda y_1 + y_2}{m + n}$)

It is to be noted that if M is the mid-point of the line segment $\overline{AB}$ , it bisects $\overline{AB}$ in 1:1 ratio. So, we have $\lambda$=1. Substituting in the above formula, we get

Mid-point of $\overline{AB}$ = ($\frac{x_1 + x_2}{2}$, $\frac{y_1 + y_2}{2}$)Hence, we proved the theorem.

### Solved Example

**Question: **Let A (-2, 3) and B(6, 7) be the mid points of a line segment $\overline{AB}$. Find the mid-point of $\overline{AB}$.

** Solution: **

Given that

A($x_1, y_1$) = (-2, 3) and B($x_2, y_2$) = (6, 7).

Then, the mid-point of $\overline{AB}$ is

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

=($\frac{-2 + 6}{2}$, $\frac{3 + 7}{2}$)

=($\frac{4}{2}$, $\frac{10}{2}$)

=(2, 5)