The Proof of the Midpoint Formula


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 $latex m$ of the line containing the two points $latex P$ and $latex Q$ with coordinates $latex (a,b)$ and $latex (c,d)$ respectively is 

$latex m = \displaystyle\frac{d – b}{c – a}$.

In addition, the distance $latex d$ between $latex PQ$ is

$latex d = \sqrt{(d-b)^2 + (c-a)^2}$.

and its midpoint $latex M$ is

$latex \displaystyle\left ( \frac{a+c}{2}, \frac{b+d}{2} \right )$.

We use these formulas to prove the midpoint formula theorem.


If segment $latex PQ$ has endpoints  $latex (a,b)$ and $latex (c,d)$, its midpoint is

$latex \displaystyle\left ( \frac{a+c}{2}, \frac{b+d}{2} \right )$.


Let $latex (a,b)$ be the coordinates of $latex P$ and $latex (c,d)$ be the coordinates of $latex Q$. In showing that $latex M$ with coordinates

$latex \displaystyle\left ( \frac{a+c}{2}, \frac{b+d}{2} \right )$

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

Midpoint Formula

For (1), we show that the distance of $latex PM$ is equal to the distance of $latex MQ$.

Distance of $latex PM$:

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

$latex |PM|= \sqrt{\left (\displaystyle\frac{a + c – 2a}{2} \right )^2 + \left( \displaystyle\frac{b + d – 2b}{2}\right )^2}$

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

Distance of $latex MQ$:

$latex |MQ| = \sqrt{\left (c – \displaystyle\frac{a+c}{2} \right)^2 + \left ( d – \displaystyle\frac{b+d}{2} \right )^2}$

$latex |MQ|= \sqrt{\left (\displaystyle\frac{2c – a – c}{2} \right )^2 + \left( \displaystyle\frac{ 2d – b – d}{2}\right )^2}$

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

Therefore, $latex PM \cong MQ$.

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

Slope of $latex PM$:

$latex \displaystyle\frac{\displaystyle\frac{b + d}{2} – b}{\displaystyle\frac{a+c}{2} – a} = \displaystyle\frac{\displaystyle\frac{b+d-2b}{2}}{\displaystyle\frac{a + c – 2a}{2}} = \displaystyle\frac{d-b}{c-a}$

Slope of $latex MQ$:

$latex \displaystyle\frac{d – \displaystyle\frac{b + d}{2}}{c – \displaystyle\frac{a+c}{2}} = \displaystyle\frac{\displaystyle\frac{2d – b – d}{2}}{\displaystyle\frac{2c – a – c}{2}} = \displaystyle\frac{d-b}{c-a}$

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

2 thoughts on “The Proof of the Midpoint Formula

  1. Pingback: Algebraic Proof: Diagonals of a Parallelogram Bisect Each Other

  2. Pingback: The Proof of the Midsegment Theorem

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