Question

If a vertex of a triangle is $(1,1)$, and the middle points of two sides passing through it are $(-2,3)$ and $(5,2)$, then find the centroid and the incenter of the triangle.

Answer 1

Let $E(-2,3)$ be the mid point of $A(1,1)$ and $B(\alpha ,\beta )$

$F(5,2)$ be the mid point of $A(1,1)$ and $C(\gamma ,\delta )$

$\frac{\alpha +1}{2}=-2\Rightarrow \alpha =-5$

$\frac{\beta +1}{2}=3\Rightarrow \beta =5$

$\frac{\gamma +1}{2}=5\Rightarrow \gamma =9$

$\frac{\delta +1}{2}=2\Rightarrow \delta =3$

$\therefore B(-5,5),C(9,3)$

$Centroid=(\frac{1-5+9}{3},\frac{1+5+3}{3})=(\frac{5}{3},3)$

$AB=\sqrt{(1+5{)}^2+(1-5{)}^2}=2\sqrt{13}$

$BC=\sqrt{(9+5{)}^2+(3-5{)}^2}=10\sqrt{2}$

$CA=\sqrt{(9-1{)}^2+(3-1{)}^2}=2\sqrt{17}$

$Incenter=(\frac{10\sqrt{2}\left(1\right)+2\sqrt{17}(-5)+2\sqrt{13}\left(9\right)}{10\sqrt{2}+2\sqrt{17}+2\sqrt{13}},\frac{10\sqrt{2}\left(1\right)+2\sqrt{17}\left(5\right)+2\sqrt{13}\left(3\right)}{10\sqrt{2}+2\sqrt{17}+2\sqrt{13}})$

$Incenter=(\frac{10\sqrt{2}-10\sqrt{17}+18\sqrt{13}}{10\sqrt{2}+2\sqrt{17}+2\sqrt{13}},\frac{10\sqrt{2}+10\sqrt{17}+6\sqrt{13}}{10\sqrt{2}+2\sqrt{17}+2\sqrt{13}})$