Question

If in a $\Delta ABC,A\equiv (1,10),$ circumcentre $\equiv (-\frac{1}{3},\frac{2}{3})$ and orthocentre $\equiv (\frac{11}{3},\frac{4}{3}),$ then the equation of side $BC$ is

• A. $12x+39y+155=0$
• B. $12x-39y+155=0$
• C. $12x+39y-155=0$
• D. $12x-39y-155=0$

Answer 1

The correct option is D $12x-39y-155=0$

Given, circumcentre, $S\equiv (-\frac{1}{3},\frac{2}{3})$
orthocentre, $H\equiv (\frac{11}{3},\frac{4}{3})$
We know that centroid divides the circumcentre and orthocentre in the ratio of $1:2$
$\therefore G\equiv (\frac{1\times \frac{11}{3}+2(-\frac{1}{3})}{1+2},\frac{1\left(\frac{4}{3}\right)+2\left(\frac{2}{3}\right)}{1+2})\Rightarrow G\equiv (1,\frac{8}{9})$

Let $D(x,y)$ is the mid-point of $B$ and $C$
$G$ divides $AD$ in $2:1$, so
$\therefore \frac{2x+1}{3}=1,\frac{2y+10}{3}=\frac{8}{9}\Rightarrow x=1,y=-\frac{11}{3}\therefore D\equiv (1,-\frac{11}{3})$

Now as we know that line $AH\perp BC$
Hence equation for side $BC$
$y+\frac{11}{3}=-\left(\frac{1-\frac{11}{3}}{10-\frac{4}{3}}\right)(x-1)\Rightarrow y+\frac{11}{3}=\frac{4}{13}(x-1)\therefore 12x-39y-155=0$