Question

The equations of the sides $AB,BC$ and $CA$ of a triangle $ABC$ are : $2x+y=0,x+py=21a,(a\ne 0)$ and $x-y=3$ respectively. Let $P(2,a)$ be the centroid of $△ABC$. Then $(BC{)}^2$ is equal to

Answer 1

Answer: 122

Assume $B(\alpha ,-2\alpha )$ and $C(\beta +3,\beta )$
$\frac{\alpha +\beta +3+1}{3}=2\text{ also }\frac{-2\alpha -2+\beta }{3}=a$
$\Rightarrow \alpha +\beta =2-2\alpha -2+\beta =3a$
$\Rightarrow \beta =2-\alpha -2\alpha -/22/2-\alpha =3a\Rightarrow \alpha =-a$
Now both $B$ and $C$ lies as given line
$\alpha -p\cdot 2\alpha =21a$
$\alpha (1-2p)=21a....$(1)
$-\alpha (1-2p)=21a\Rightarrow p=11$
$\beta +3+p\beta =21a$
$\beta +3+11\beta =21a$
$21\alpha +12\beta +3=0$
Also $\beta =2-\alpha$
$21\alpha +12(2-\alpha )+3=0$
$21\alpha +24-12\alpha +3=0$
$9\alpha +27=0$
$\alpha =-3,\beta =5$
So $BC=\sqrt{122}$ and $(BC{)}^2=122$