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The equations of two sides $\mathrm{AB}$ and $\mathrm{AC}$ of a triangle $\mathrm{ABC}$ are $4 x+y=14$ and $3 x-2 y=5$, respectively. The point $\left(2,-\frac{4}{3}\right)$ divides the third side $\mathrm{BC}$ internally in the ratio $2: 1$. the equation of the side $\mathrm{BC}$ is
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Verified Answer
The correct answer is:
$x+3 y+2=0$
$\begin{aligned}
& \frac{2 \mathrm{x}_2+\mathrm{x}_1}{3}=2, \frac{2\left(\frac{3 \mathrm{x}_2-5}{2}\right)+\left(14-4 \mathrm{x}_1\right)}{3}=\frac{-4}{3} \\
& 2 \mathrm{x}_2+\mathrm{x}_1=6,3 \mathrm{x}_2-4 \mathrm{x}_1=-13 \\
& \mathrm{x}_2=1, \mathrm{x}_1=4
\end{aligned}$
So, $\mathrm{C}(1,-1), \mathrm{B}(4,-2)$
$\mathrm{m}=\frac{-1}{3}$
Equation of $\mathrm{BC}: \mathrm{y}+1=\frac{-1}{3}(\mathrm{x}-1)$
$\begin{aligned} & 3 y+3=-x+1 \\ & x+3 y+2=0\end{aligned}$
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