Question

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

Answer 1

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=3 a$
$ \Rightarrow \beta=2-\alpha \,\,\,\,-2 \alpha-\not 22 \not \not 2-\alpha=3 a \Rightarrow \alpha=- a$
Now both $B$ and $C$ lies as given line
$ \alpha- p \cdot 2 \alpha=21 a$
$ \alpha(1-2 p )=21 a....$(1)
$ -\alpha(1-2 p )=21 a \Rightarrow p =11 $
$\beta+3+ p \beta=21 a$
$ \beta+3+11 \beta=21 a $
$ 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 $( BC )^2=122$