Question

If a triangle $ABC,A\equiv (1,10)$, circumcenter $\equiv (-\frac{1}{3},\frac{2}{3})$, orthocentre $\equiv (\frac{11}{3},\frac{4}{3})$ then the coordinate mid-point of side opposite to $A$ is?

Answer 1

Let $x_1,x_2,x_3$ be the $x-coordinates.$ and $y_1,y_2,y_3$ be the $y-coordinates$ of $\Delta ABC$.

As, $O,G,C$ are standard notations for orthocenter,centroid,circumcenter respectively.

Let $(X,Y)$ be coordinates of centroid.

To Find: Midpoint of BC = $(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})$

Since we know , $G$ divides line joining $OGC$ in ratio $2:1$ resp.

so,

$X=\frac{2\left(\frac{-1}{3}\right)+1\left(\frac{11}{3}\right)}{3}$

$\Rightarrow X=1$

$Y=\frac{2\left(\frac{2}{3}\right)+1\left(\frac{4}{3}\right)}{3}$

$\Rightarrow Y=\frac{8}{9}$

Now as,

$X=\frac{x_1+x_2+x_3}{3}$

on solving we get ,

$\frac{x_2+x_3}{2}=1$

Similarly we can find,

$\frac{y_2+y_3}{2}=\frac{-11}{3}$

Thus our answer is = $(1,\frac{-11}{3})$