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.

Midpoint

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?).

Midpoint Formula

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.

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29. December 2012 by Guillermo Bautista
Categories: Algebra, Geometry | Tags: , , , , , | 2 comments