Answer
See below
Work Step by Step
Let's note $A(x_1,y_1)$, $B(x_2,y_2)$.
Determine the distance $d_{AM}$ and the distance $d_{BM}$ using the distance formula:
$$\begin{align*}
d_{AM}&=\sqrt{(x_A-x_M)^2+(y_A-y_M)^2}\\
&=\sqrt{\left(x_1-\dfrac{x_1+x_2}{2}\right)^2+\left(y_1-\dfrac{y_1+y_2}{2}\right)^2}\\
&=\sqrt{\dfrac{(x_1-x_2)^2}{4}+\dfrac{(y_1-y_2)^2}{4}}\\
&=\dfrac{1}{2}\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\
&=\dfrac{1}{2}d_{AB}.
\end{align*}$$
$$\begin{align*}
d_{BM}&=\sqrt{(x_B-x_M)^2+(y_B-y_M)^2}\\
&=\sqrt{\left(x_2-\dfrac{x_1+x_2}{2}\right)^2+\left(y_2-\dfrac{y_1+y_2}{2}\right)^2}\\
&=\sqrt{\dfrac{(x_2-x_1)^2}{4}+\dfrac{(y_2-y_1)^2}{4}}\\
&=\dfrac{1}{2}\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\
&=\dfrac{1}{2}d_{AB}.
\end{align*}$$
Because $d_{AM}=d_{BM}$, we got that $M$ is equidistant from each endpoint $A$ and $B$.
Because $d_{AM}=d_{BM}=\frac{1}{2}d_{AB}$, it means that $M$ lies on the segment $\overline{AB}$ and is its midpoint.
