Question

Find the co-ordinates of the incentre of the triangle whose vertices are $(-2,4),(5,5)$ and $(4,-2)$

Answer 1

Let $A(-2,4),B(5,5)$ and $C(4,-2)$ be the vertices of the given $△ABC$ then

$a=\left|BC\right|=\sqrt{{(4-5)}^2+{(-2-5)}^2}=\sqrt{1+49}=\sqrt{50}=5\sqrt{2}$

$b=\left|CA\right|=\sqrt{{(4+2)}^2+{(-2-4)}^2}=\sqrt{36+36}=\sqrt{36\times 2}=6\sqrt{2}$

and $c=\left|AB\right|=\sqrt{{(5+2)}^2+{(5-4)}^2}=\sqrt{49+1}=\sqrt{50}=5\sqrt{2}$

$\therefore$ The coordinates of the incentre of $△ABC$ is

$(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$

$=(\frac{5\sqrt{2}\times -2+6\sqrt{2}\times 5+5\sqrt{2}\times 4}{5\sqrt{2}+6\sqrt{2}+5\sqrt{2}},\frac{5\sqrt{2}\times 4+6\sqrt{2}\times 5+5\sqrt{2}\times -2}{5\sqrt{2}+6\sqrt{2}+5\sqrt{2}})$

$=(\frac{-10\sqrt{2}+30\sqrt{2}+20\sqrt{2}}{16\sqrt{2}},\frac{20\sqrt{2}+30\sqrt{2}-10\sqrt{2}}{16\sqrt{2}})$

$=(\frac{40\sqrt{2}}{16\sqrt{2}},\frac{40\sqrt{2}}{16\sqrt{2}})$

$=(\frac{5}{2},\frac{5}{2})$