A triangle has corners at (2 , 1 ), (1 ,3 ), and (3 ,4 ). What is the radius of the triangle's inscribed circle?

Mar 29, 2018

$\frac{\sqrt{5} \left(2 - \sqrt{2}\right)}{2}$.

Explanation:

Let the vertices of $\Delta A B C$ be $A \left(2 , 1\right) , B \left(1 , 3\right) \mathmr{and} C \left(3 , 4\right)$, and

let $r$ be the inradius of $\Delta A B C$.

Then, ${a}^{2} = B {C}^{2} = {\left(1 - 3\right)}^{2} + {\left(3 - 4\right)}^{2} = 4 + 1 = 5$,

${b}^{2} = C {A}^{2} = {\left(3 - 2\right)}^{2} + {\left(4 - 1\right)}^{2} = 1 + 9 = 10$, and,

${c}^{2} = A {B}^{2} = {\left(2 - 1\right)}^{2} + {\left(1 - 3\right)}^{2} = 1 + 4 = 5$.

$\therefore {c}^{2} + {a}^{2} = {b}^{2} \Rightarrow \angle B = {90}^{\circ}$.

$\therefore \Delta A B C$ is a right-triangle.

We know from Geometry, then,

$r = \frac{c + a - b}{2} = \frac{\sqrt{5} + \sqrt{5} - \sqrt{10}}{2} = \frac{\sqrt{5} \left(2 - \sqrt{2}\right)}{2}$.