Question

Let in a $\Delta ABC,$ vertex $A,$ the circumcentre $S$ and the orthocentre $H$ are $\left(1,10\right),\left(-\frac{1}{3},\frac{2}{3}\right)$ and $\left(\frac{11}{3},\frac{4}{3}\right)$ respectively, then mid-point of the side $BC$ is

• A. $\left(1,\frac{-11}{3}\right)$
• B. $\left(1,\frac{-22}{3}\right)$
• C. $\left(2,\frac{-11}{3}\right)$
• D. $\left(-1,\frac{11}{3}\right)$

Answer 1

Answer: (A)

We know that centroid divides orthocentre and cirumcentre in the ratio $2:1$
Hence, centroid $\left(G\right)$ is $\left(\frac{-\frac{2}{3}+\frac{11}{3}}{3},\frac{\frac{4}{3}+\frac{4}{3}}{3}\right)$
$\Rightarrow G=\left(1,\frac{8}{9}\right)$
Let, $D$ is the mid-point of side $BC$ then $G$ also divides $A$ and $D$ in the ratio $2:1$

$\frac{2h+1}{3}=1$ and $\frac{2k+10}{3}=\frac{8}{9}$
$\Rightarrow h=1$ and $k=-\frac{11}{3}\Rightarrow D=\left(1,-\frac{11}{3}\right)$