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