Question

In a $△ABC,$ the coordinates of $A$ are $(1,10).$ If the coordinates of orthocentre and circumcentre are $(\frac{11}{3},\frac{4}{3})$ and $(-\frac{1}{3},\frac{2}{3}),$ then the side $BC$ passes through the point

• A. $(-\frac{31}{3},-\frac{29}{39})$
• B. $(-\frac{31}{3},\frac{29}{39})$
• C. $(\frac{31}{3},\frac{31}{39})$
• D. $(\frac{31}{3},-\frac{31}{39})$

Answer 1

The correct option is D $(\frac{31}{3},-\frac{31}{39})$

Let $G$ be the centroid of $△ABC$
and $H$ and $O$ be the orthocentre and circumcentre respectively.
We know that the centroid divides the distance from the orthocentre to the circumcentre in the ratio $2:1$
i.e., $H+2O=3G$
$\therefore G\equiv (1,\frac{8}{9})$

Let $D$ be the mid-point of side $BC.$
Then $AG:GD=2:1$
$\Rightarrow D\equiv (1,-\frac{11}{3})$
Slope of $AH$ is $\frac{10-4/3}{1-11/3}=-\frac{13}{4}$
$BC\perp AH$
$\therefore$ Slope of $BC=\frac{4}{13}$
Equation of $BC$ is $y+\frac{11}{3}=\frac{4}{13}(x-1)$
or, $12x-39y-155=0$
From options, $BC$ passes through $(\frac{31}{3},-\frac{31}{39})$