The equations of two sides [AB] and [AC] of a triangle [ABC] are 4 x+y=14 and 3 x-2 y=5, r…

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

Accepted Answer

$$\begin{aligned}
& 2=\frac{2 x_2+x_1}{3}, \frac{-4}{3}=\frac{2 y_2+y_1}{3} \
& 2 x_2+x_1=6,2 y_2+y_1=-4
\end{aligned}$$

$$\begin{aligned}
& x_1=6-2 x_2 \quad \text{.... (1)}\
& y_1=-4-2 y_2 \quad \text{.... (2)}\
& 4 x_1+y_1=14 \quad \text{.... (3)}\
& 3 x_2-2 y_2=5 \quad \text{.... (4)}
\end{aligned}$$

From here, $x_2=1, y_2=-1, x_1=4, y_1=-2$

$$\begin{aligned}
& B(4,-2) C(1,-1) \
& y+2=\frac{-1+2}{1-4}(x-4) \
& -3 y-6=x-4 \
& x+3 y+2=0
\end{aligned}$$

Solutions · 1