# The Proof of the Midpoint Formula

Introduction

The midpoint is the middle point of a segment. It is equidistant from both end points. The midpoint formula is used to determine the coordinates of the midpoint of a segment. This formula is familiar to middle school and high school students, however most books do not discuss its proof. In this post, we discuss the mathematical proof of the midpoint formula. The slope and distance formulas are needed to prove the midpoint formula. Recall that the slope $m$ of the line containing the two points $P$ and $Q$ with coordinates $(a,b)$ and $(c,d)$ respectively is $m = \displaystyle\frac{d - b}{c - a}$.

In addition, the distance $d$ between $PQ$ is $d = \sqrt{(d-b)^2 + (c-a)^2}$.

and its midpoint $M$ is $\displaystyle\left ( \frac{a+c}{2}, \frac{b+d}{2} \right )$.

We use these formulas to prove the midpoint formula theorem.

Theorem

If segment $PQ$ has endpoints $(a,b)$ and $(c,d)$, its midpoint is $\displaystyle\left ( \frac{a+c}{2}, \frac{b+d}{2} \right )$.

Proof

Let $(a,b)$ be the coordinates of $P$ and $(c,d)$ be the coordinates of $Q$. In showing that $M$ with coordinates $\displaystyle\left ( \frac{a+c}{2}, \frac{b+d}{2} \right )$

is the midpoint of $PQ$, we have to show that (1) $PM \cong MQ$ and (2) $M$ is on $PQ$ (Why?). For (1), we show that the distance of $PM$ is equal to the distance of $MQ$.

Distance of $PM$: $|PM| = \sqrt{\left (\displaystyle\frac{a+c}{2}-a \right)^2 + \left (\displaystyle\frac{b+d}{2}-b \right )^2}$ $|PM|= \sqrt{\left (\displaystyle\frac{a + c - 2a}{2} \right )^2 + \left( \displaystyle\frac{b + d - 2b}{2}\right )^2}$ $|PM|= \sqrt{\left ( \displaystyle\frac{c-a}{2} \right )^2 + \left ( \displaystyle\frac{d-b}{2} \right )^2 }$

Distance of $MQ$: $|MQ| = \sqrt{\left (c - \displaystyle\frac{a+c}{2} \right)^2 + \left ( d - \displaystyle\frac{b+d}{2} \right )^2}$ $|MQ|= \sqrt{\left (\displaystyle\frac{2c - a - c}{2} \right )^2 + \left( \displaystyle\frac{ 2d - b - d}{2}\right )^2}$ $|MQ|= \sqrt{\left ( \displaystyle\frac{c-a}{2} \right )^2 + \left ( \displaystyle\frac{d-b}{2} \right )^2 }$

Therefore, $PM \cong MQ$.

For (2), we have to show that the slope of $PM$ is equal to the slope of $MQ$.

Slope of $PM$: $\displaystyle\frac{\displaystyle\frac{b + d}{2} - b}{\displaystyle\frac{a+c}{2} - a} = \displaystyle\frac{\displaystyle\frac{b+d-2b}{2}}{\displaystyle\frac{a + c - 2a}{2}} = \displaystyle\frac{d-b}{c-a}$

Slope of $MQ$: $\displaystyle\frac{d - \displaystyle\frac{b + d}{2}}{c - \displaystyle\frac{a+c}{2}} = \displaystyle\frac{\displaystyle\frac{2d - b - d}{2}}{\displaystyle\frac{2c - a - c}{2}} = \displaystyle\frac{d-b}{c-a}$

The slope of $PM$ and $MQ$ are equal and $PM \cong MQ$, therefore $M$ is the midpoint of $PQ$.