The equations of the sides AB,BC,and CA of a | vedclass

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(D) The vertices are the intersection points of the lines.$A$ is the intersection of $2x + y = 0$ and $x - y = 3$. Adding these,$3x = 3 \Rightarrow x

Vedclass · Accepted Answer

$122$

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(D) The vertices are the intersection points of the lines.$A$ is the intersection of $2x + y = 0$ and $x - y = 3$. Adding these,$3x = 3 \Rightarrow x = 1$. Then $y = -2$. So $A = (1, -2)$.$B$ is the intersection of $2x + y = 0$ and $x + py = 21a$. Let $B = (\alpha, -2\alpha)$.$C$ is the intersection of $x - y = 3$ and $x + py = 21a$. Let $C = (\beta + 3, \beta)$.The centroid $G(2, a) = (\frac{1 + \alpha + \beta + 3}{3}, \frac{-2 - 2\alpha + \beta}{3})$.Equating coordinates: $1 + \alpha + \beta + 3 = 6$ $\Rightarrow \alpha + \beta = 2$ $\Rightarrow \beta = 2 - \alpha$.$-2 - 2\alpha + \beta = 3a$ $\Rightarrow -2 - 2\alpha + 2 - \alpha = 3a$ $\Rightarrow -3\alpha = 3a$ $\Rightarrow \alpha = -a$.Since $B$ lies on $x + py = 21a$: $\alpha + p(-2\alpha) = 21a$ $\Rightarrow \alpha(1 - 2p) = 21(- \alpha)$ $\Rightarrow 1 - 2p = -21$ $\Rightarrow 2p = 22$ $\Rightarrow p = 11$.Since $C$ lies on $x + py = 21a$: $(\beta + 3) + 11\beta = 21a$ $\Rightarrow 12\beta + 3 = 21(- \alpha)$ $\Rightarrow 4\beta + 1 = -7\alpha$.Substitute $\beta = 2 - \alpha$: $4(2 - \alpha) + 1 = -7\alpha$ $\Rightarrow 8 - 4\alpha + 1 = -7\alpha$ $\Rightarrow 3\alpha = -9$ $\Rightarrow \alpha = -3$.Then $\beta = 2 - (-3) = 5$. Thus $B = (-3, 6)$ and $C = (8, 5)$.$(BC)^2 = (8 - (-3))^2 + (5 - 6)^2 = 11^2 + (-1)^2 = 121 + 1 = 122$.

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

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