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

• Correct answer
  A

  $x+3 y+2=0$
• B

  $x-6 y-10=0$
• C

  $x-3 y-6=0$
• D

  $x+6 y+6=0$

Solution

$\begin{aligned} & \frac{2 \mathrm{x}_2+\mathrm{x}_1}{3}=2, \frac{2\left(\frac{3 \mathrm{x}_2-5}{2}\right)+\left(14-4 \mathrm{x}_1\right)}{3}=\frac{-4}{3} \\ & 2 \mathrm{x}_2+\mathrm{x}_1=6,3 \mathrm{x}_2-4 \mathrm{x}_1=-13 \\ & \mathrm{x}_2=1, \mathrm{x}_1=4 \end{aligned}$
So, $\mathrm{C}(1,-1), \mathrm{B}(4,-2)$ $\mathrm{m}=\frac{-1}{3}$ Equation of $\mathrm{BC}: \mathrm{y}+1=\frac{-1}{3}(\mathrm{x}-1)$ $\begin{aligned} & 3 y+3=-x+1 \\ & x+3 y+2=0\end{aligned}$

Hence, the correct answer is option: 'A'.