Question

If in triangle ABC, $A\equiv (1,10),\text{circumcenter}\equiv (-\frac{1}{3},\frac{2}{3})$and orthocenter $\equiv (\frac{11}{3},\frac{4}{3})$,then the coordinates of mid-point of side opposite to A is

Answer 1

In triangle ABC, $A\equiv (1,10)$, circumcenter $\left(0\right)\equiv (-\frac{1}{3},\frac{2}{3})$ and orthocenter $\left(H\right)\equiv (\frac{11}{3},\frac{4}{3})$

Therefore the coordinates of $G(1,\frac{8}{9})$
As AD : DG =3 :1
$D_x=\frac{3-1}{2}=1,D_y=\frac{(\frac{8}{3}-10)}{2}=-\frac{11}{3}$
Hence the coordinates of mid point of BC is $(1,-\frac{11}{3})$