JEE MAIN - Mathematics (2023 - 24th January Evening Shift - No. 19)
The equations of the sides AB, BC and CA of a triangle ABC are : $$2x+y=0,x+py=21a,(a\pm0)$$ and $$x-y=3$$ respectively. Let P(2, a) be the centroid of $$\Delta$$ABC. Then (BC)$$^2$$ is equal to ___________.
Answer
122
Explanation
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$$ \begin{aligned} & \therefore 4 p^{2}-21 a p+8 p+42 a-5=0\quad...(1) \end{aligned} $$
And $\frac{-42 a}{1-2 p}-2+\frac{21 a-3}{p+1}=3 a$
$$ \therefore 4 p^{2}-81 a p+6 a p^{2}-24 a+8 p-5=0 \quad...(2) $$
From equation (1) - equation (2) we get;
$$ 60 a p+66 a-6 a p^{2}=0 $$
$$ \begin{aligned} \because a \neq 0 \Rightarrow p^{2}-10 p-11=0 \\\\ p=-1 \text { or } 11 \Rightarrow p=11 . \end{aligned} $$
When $p=11$ then $a=3$
Coordinate of $B=(-3,6)$
And coordinate of $C=(8,5)$
$\therefore B C^{2}=122$
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