Why does averaging the coordinates of the endpoints of a line segment give you the midpoint? Why does taking the average of both the $x$ and $y$ values of the endpoints of a line segment in $\Bbb R^2$ result in the coordinates of the midpoint?
 A: There are two ways to visualise it:
First, coordinate geometry. $x$ and $y$ are orthogonal, and therefore independent (a line parallel to the $x$-axis has a null projection on the $y$-axis, and vice versa). This means that they can be treated independently. So intuitively, you can find the average of the $x$ and $y$ values of the endpoints independently and that will give you the midpoint of the segment.
Second, and more rigorously, vectors. If you denote the vector along the line segment as $(x_2 - x_1)\vec{i} + (y_2 - y_1)\vec{j}$, then the vector in the direction from $(x_1,y_1)$ to $(x_2, y_2)$ that denotes half the line segment is $\frac 12 ((x_2 - x_1)\vec{i} + (y_2 - y_1)\vec{j})$. This vector extends from $(x_1,y_1)$ to the midpoint of the line segment. To find the coordinate vector for the midpoint, you need to add the coordinate vector for $(x_1,y_1)$ to this, and that gives you $x_1 \vec{i} + y_1 \vec{j} + \frac 12 ((x_2 - x_1)\vec{i} + (y_2 - y_1)\vec{j}) = \frac 12((x_2 + x_1)\vec{i} + (y_2 + y_1)\vec{j})$, which immediately gives you the coordinate of the midpoint as $(\frac{x_1 + x_2}{2}), (\frac{y_1 + y_2}{2})$.
A: Hint: Let the two points be $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$. Then the distance between these two points is $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$
Thus if $M=(x_0,y_0)$ and is the midpoint of $P_1P_2$ then
$$P_1M=P_2M=\frac{1}{2}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$
