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

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 $∆ A B C$ . Then ${(B C)}^{2}$ is equal to

Smart Achievers · Accepted Answer

(4)

$2 x + y = 0 ------(1)$

$x - y = 0 ------(2)$

$x + p y = 21 a ------(3)$

$Solving (1) & (2) \Rightarrow A (1, - 2)$

$Centroid of triangle A B C is (\frac{4 + s + t}{3}, \frac{- 2 - 2 s + t}{3}) = (2, a)$

$\Rightarrow s + t = 2 ------(4)$

$\Rightarrow s = - a, t = 2 + a$

$- 2 s + t = 3 a + 2 ------(5)$

$Solving (4) & (5) we get$

$B (- a, 2 a), C (a + 5, a + 2)$

$∴$ $Distance {(B C)}^{2} = 122$

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