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

The correct answer is 122

Assume B(α,−2α) and C(β+3,β)
3α+β+3+1​=2 also 3−2α−2+β​=a
⇒α+β=2−2α−2+β=3a
⇒β=2−α−2α−222−α=3a⇒α=−a
Now both B and C lies as given line
α−p⋅2α=21a
α(1−2p)=21a....(1)
−α(1−2p)=21a⇒p=11
β+3+pβ=21a
β+3+11β=21a
21α+12β+3=0
Also β=2−α
21α+12(2−α)+3=0
21α+24−12α+3=0
9α+27=0
α=−3,β=5
So BC=122​ and (BC)2=122