This is Question 43 from Stewart's review of analytic geometry, which can be found in PDF form here: https://www.stewartcalculus.com/data/CALCULUS_8E_ET/upfiles/6et_reviewofanalgeom.pdf
There is no answer supplied for this question so I was wondering if someone might review my answer for errors, or suggest any alternative approaches?
The question asks you to show that the midpoint of a line segment from $P_1(x_1,y_1)$ to $P_2(x_2,y_2)$ is:
$$\left(\frac{(x_1+x_2)}{2},\frac{(y_1+y_2)}{2}\right)$$
My solution is as follows:
Take $P_1$ and $P_2$ to be points sitting opposite each other on a circle, so that the line segment that connects them has length $D$, the diameter of the circle. Then the centre of the circle $C(h, k)$ would coincide with the midpoint of the line segment, and the radius of the circle is $r=D/2$.
Taking $\Delta x=x_2-x_1$ and $\Delta y=y_2-y_1$, by Pythagoras's theorem,
$$D^2=(x_2-x_1)^2+(y_2-y_1)^2$$
Then,
$$r^2=\frac{D^2}{2^2}=\frac{(x_2-x_1)^2}{4}+\frac{(y_2-y_1)^2}{4}$$
The equation of a circle is:
$$r = \sqrt{(x-h)^2+(y-k)^2}$$
Squaring the RHS, choosing the point on the circle, $P_1(x_1,y_1)$, and substituting out $h$ and $k$,
$$\begin{align}(x-h)^2+(y-k)^2&=\left(x_1-\frac{x_1+x_2}{2}\right)^2+\left(y_1-\frac{y_1+y_2}{2}\right)^2\\&=\left(\frac{x_1-x_2}{2}\right)^2+\left(\frac{y_1-y_2}{2}\right)^2\\&=\frac{(x_2-x_1)^2}{4}+\frac{(y_2-y_1)^2}{4}\\&=r^2\end{align}$$
Which brings us back to the equation for $r^2$ above.
The midpoint is $$ P_1+\frac{P_1P_2}{2}=(x_1,y_1)+\frac{1}{2}(x_2-x_1,y_2-y_1) $$ $$ =(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}). $$ This is a purely affine fact. You do not need to invoke distances/radii/circles.