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

Question

The equations of two sides $AB$ and $AC$ of a triangle $ABC$ are $4x+y=14$ and $3x-2y=5$, respectively. The point $(2,-\frac{4}{3})$ divides the third side $BC$ internally in the ratio $2:1$, the equation of the side $BC$ is

Accepted Answer

$\begin{gathered} 2=\frac{2{x}_{2}+{x}_{1}}{3},\frac{-4}{3}=\frac{2{y}_{2}+{y}_{1}}{3} \ 2{x}_{2}+{x}_{1}=6,2{y}_{2}+{y}_{1}=-4 \end{gathered}$

$\begin{gathered} {x}_{1}=6-2{x}_{2}\, \ {y}_{1}=-4-2{y}_{2}\, \ 4{x}_{1}+{y}_{1}=14\, \ 3{x}_{2}-2{y}_{2}=5\, \end{gathered}$

From here, ${x}_{2}=1,{y}_{2}=-1,{x}_{1}=4,{y}_{1}=-2$

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

Solutions · 1