JEE MAIN - Mathematics (2024 - 8th April Morning Shift - No. 15)
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
$$x+6 y+6=0$$
$$x-3 y-6=0$$
$$x+3 y+2=0$$
$$x-6 y-10=0$$
Explanation
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$$\begin{aligned} & 2=\frac{2 x_2+x_1}{3}, \frac{-4}{3}=\frac{2 y_2+y_1}{3} \\ & 2 x_2+x_1=6,2 y_2+y_1=-4 \end{aligned}$$
$$\begin{aligned} & x_1=6-2 x_2 \quad \text{.... (1)}\\ & y_1=-4-2 y_2 \quad \text{.... (2)}\\ & 4 x_1+y_1=14 \quad \text{.... (3)}\\ & 3 x_2-2 y_2=5 \quad \text{.... (4)} \end{aligned}$$
From here, $$x_2=1, y_2=-1, x_1=4, y_1=-2$$
$$\begin{aligned} & B(4,-2) C(1,-1) \\ & y+2=\frac{-1+2}{1-4}(x-4) \\ & -3 y-6=x-4 \\ & x+3 y+2=0 \end{aligned}$$
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