I have three points
$A(-10,-12)$,
$B(6,18)$,
$C(-2,-14)$.
I have to prove they exist on the same semicircle. How?

If I try to draw the three points to investigate I see that I can draw a right triangle. I know you can inscribe a right triangle between three points of a semicircle, the diameter being the hypotenuse of this triangle.

So I thought I could do Pythagoras/distance formula between the points to find $AB$, $BC$, and $AC$. Then verify with Pythagoras whether $\sqrt{AC^2+BC^2}=BA$ , but for some reason this does not add up to the expected result (why not?)

So I looked up the solution and it reads:
\begin{align\*}
m\_{AB} &= \frac{y\_B-y\_A}{x\_B-x\_A} 
= \phantom{-}\frac{18+12}{6+10} = \frac{30}{16} = \frac{15}{8} \\
m\_{BC} &= \frac{y\_C-y\_B}{x\_C-x\_B} 
= \frac{-14-18}{-2-6} = \frac{-32}{-8} = 4 \\
m\_{CA} &= \frac{y\_A-y\_C}{x\_A-x\_C} 
= \frac{-12+14}{-10+2} = \frac{2}{-8} = -\frac{1}{4}
\end{align\*}

Since
$$
4 \cdot \frac{-1}{4} = -1,
$$
$BC$ is perpendicular to $CA$ (negative reciprocal) and hence $\triangle ABC$ is a right angled triangle.

Since the angle in a semicircle is a right angle it follows that $AB$ is a diameter and the points form a semicircle.

So I understand that

> 
> $$m\_{AB} = \frac{y\_B- y\_A}{x\_B-x\_A}$$
> 
> 
> 

Is the gradient formula
$$
\frac{\text{rise}}{\text{run}}
$$

So basically this (more efficient) solution is ignoring lengths/distances and leapfrogging lines and is going straight to prove there is a right angle (clever!)

But I wonder... why would my solution still not work?