Incircle Radius of Triangle

Let vertices be $A(1,2)$, $B(7,2)$, and $C(1,10)$.

• $AC$ is vertical of length $|10 - 2| = 8$.
• $AB$ is horizontal of length $|7 - 1| = 6$. Since $AC$ is vertical and $AB$ is horizontal, $\angle A = 90^\circ$.

So $\triangle ABC$ is a right-angled triangle with perpendicular sides $a = 6$ and $b = 8$. Hypotenuse $c = \sqrt{6^2 + 8^2} = 10$.

Inradius $r = \frac{a + b - c}{2} = \frac{6 + 8 - 10}{2} = 2$.