Question

The x-coordinates of the incentre of the triangle that has the coordinates of the midpoint of its sides as $(0,1),(1,1),$ and $(1,0)$ is $\frac{a}{2+\sqrt{2}}$. Find $a$.

Answer 1

$a=\sqrt{{(1-1)}^2+{(1-0)}^2}=1b=\sqrt{{(1-0)}^2+{(0-1)}^2}=\sqrt{2}c=\sqrt{{(1-0)}^2+{(1-1)}^2}=1$

x-coordinate of incentre=$\frac{(a\times 0)+(b\times 1)+(c\times 1)}{a+b+c}=\frac{0+\sqrt{2}+1}{1+\sqrt{2}+1}=\frac{\sqrt{2}+1}{2+\sqrt{2}}\therefore a=\sqrt{2}+1$