The equations of two sides AB and AC of a tri | vedclass

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(C) Let $B$ be $(x_1, 14 - 4x_1)$ and $C$ be $(x_2, \frac{3x_2 - 5}{2})$.Given that the point $P\left(2, -\frac{4}{3}\right)$ divides $BC$ in the rati

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$x + 3y + 2 = 0$

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(C) Let $B$ be $(x_1, 14 - 4x_1)$ and $C$ be $(x_2, \frac{3x_2 - 5}{2})$.Given that the point $P\left(2, -\frac{4}{3}\right)$ divides $BC$ in the ratio $2:1$,we use the section formula:$2 = \frac{2x_2 + x_1}{2 + 1} \implies 2x_2 + x_1 = 6$ (Equation $1$)$-\frac{4}{3} = \frac{2\left(\frac{3x_2 - 5}{2}\right) + (14 - 4x_1)}{3} \implies -4 = 3x_2 - 5 + 14 - 4x_1 \implies 3x_2 - 4x_1 = -13$ (Equation $2$)Solving Equations $1$ and $2$:Multiply Equation $1$ by $4$: $8x_2 + 4x_1 = 24$Adding this to Equation $2$: $11x_2 = 11 \implies x_2 = 1$Substituting $x_2 = 1$ in Equation $1$: $2(1) + x_1 = 6 \implies x_1 = 4$Thus,$B = (4, 14 - 4(4)) = (4, -2)$ and $C = (1, \frac{3(1) - 5}{2}) = (1, -1)$.The slope of $BC$ is $m = \frac{-1 - (-2)}{1 - 4} = \frac{1}{-3} = -\frac{1}{3}$.The equation of $BC$ is $y - (-1) = -\frac{1}{3}(x - 1) \implies 3y + 3 = -x + 1 \implies x + 3y + 2 = 0$.

The equations of two sides $AB$ and $AC$ of a triangle $ABC$ are $4x + y = 14$ and $3x - 2y = 5$,respectively. The point $\left(2, -\frac{4}{3}\right)$ divides the third side $BC$ internally in the ratio $2:1$. The equation of the side $BC$ is:

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