JEE Math Practice Question
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
- A.$$x+6 y+6=0$$
- B.$$x-3 y-6=0$$
- C.$$x+3 y+2=0$$
- D.$$x-6 y-10=0$$
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