Coordinate geometry, proof of midpoint

Question

I am reviewing my knowledge of coordinate geometry and came across this question about proving the midpoint formula:

Unfortunately, the answer for this question just says ‘proof’.

I am fairly confident with my answer (posted below) but I found it to be quite long, so just wondering if there is a quicker method to prove this (also if someone could quickly read through my proof to make sure it is correct, I would be very happy, thanks).

Answer 1

Alternatively, you can use geometrical approach. Refer to the figure:

$\hspace{3cm}$

The points $A(x_1,y_1),B(x_2,y_2),M\left(x_m=\frac{x_1+x_2}{2},y_m=\frac{y_1+y_2}{2}\right)$ are given.

First note that the points $A$, $M$ and $B$ are colinear (i.e. lie on the same line): $$\text{slope of AM}=\frac{y_m-y_1}{x_m-x_1}=\frac{\frac{y_1+y_2}{2}-y_1}{\frac{x_1+x_2}{2}-x_1}=\frac{y_2-y_1}{x_2-x_1}=\text{slope of AB}.$$

The triangles $\Delta AMD$ and $\Delta ABC$ are similar, because their corresponding angles are equal. Hence: $$\frac{AM}{AB}=\frac{AD}{AC} =\frac{AD}{2AD}=\frac12 \Rightarrow AM=\frac12AB.$$

Answer 2

A slightly shorter way would be\begin{align}AM^2&=\left(\frac{x_1+x_2}2-x_1\right)^2+\left(\frac{y_1+y_2}2-y_1\right)^2\\&=\left(\frac{-x_1+x_2}2\right)^2+\left(\frac{-y_1+y_2}2\right)^2\\&=\frac14\bigl((-x_1+x_2)^2+(-y_1+y_2)^2\bigr)=\frac14AB^2,\end{align}from which it follows that $AM=\frac12AB$.

Answer 3

You may say it this way:

The point $P=\Big(tx_1+[1-t]x_2,ty_1+[1-t]y_2\Big)$ falls within the line section between $(x_1,y_1)$ and $(x_2,y_2)$ for $0<t<1$. The distance from $(x_1,y_1)$ to $P$ is proportional to $t$ (which can be easily proved using the distance formula). Here for the midpoint we have $t={1\over 2}$ and hence the result.