The equations of the sides AB, BC and CA of a triangle ABC are: 2 x + y = 0 , x + p y = 21 a ( a 0 ) and x - y = 3 respectively. Let P ( 2 , a ) be the centroid of A B C . Then ( B C ) 2 is equal to _______ . [2023]

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Solution

Q.
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 $∆ A B C$ . Then ${(B C)}^{2}$ is equal to _______ . [2023]

Ans.
(122)

$2 = \frac{1 + α + β + 3}{3}$

$\Rightarrow 6 = 4 + α + β$

$\Rightarrow α + β = 2$

$∴$ $β = 2 - α . . . (i)$

$∴$ $C (5 - α, 2 - α)$

$Now, \frac{- 2 - 2 α + β}{3} = a$

$\Rightarrow - 2 - 2 α + β = 3 a \Rightarrow β - 2 α = 3 a + 2$

$\Rightarrow 2 - α - 2 α = 3 a + 2$

$∴$ $α = - a \Rightarrow β = 2 + a$

$Point B, C lies on x + p y = 21 a$

$So, α - 2 α p = 21 a \Rightarrow - a + 2 a p = 21 a \Rightarrow 2 a p = 22 a$

$\Rightarrow p a = 11 a . . . (i i)$

$∴$ $p = 11, a = 0 (rejected)$

$and β + 3 + p β = 21 a$

$\Rightarrow 5 + a + p (2 + a) = 21 a \Rightarrow 5 + a + 2 p + p a = 21 a$

$\Rightarrow 5 + a + 2 p + 11 a = 21 a \Rightarrow 2 p + 5 = 21 a - 12 a$

$\Rightarrow 2 p + 5 = 9 a \Rightarrow 2 \times 11 + 5 = 9 a$ $∴ a = 3$

$∴$ Point $B$ is $(- 3, 6)$ and $C (8, 5)$

$∴$ $B C = \sqrt{{(8 + 3)}^{2} + {(5 - 6)}^{2}} \Rightarrow {(B C)}^{2} = 122$

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