Question

The sides of a triangle $ABC$ are as under $:AB:2x+3y=29$ and $AC:x+2y=16$. If the mid-point of $BC$ is $(5,6)$ then find the equation of $BC$.

Answer 1

The line $BC$ is $(y-6)=m(x-5)$
$\frac{x-5}{\cos \theta }=\frac{y-6}{\sin \theta }=\stackrel{r_{,}}{C}\stackrel{-r}{B}$
$\therefore C(r\cos \theta +5,r\sin \theta +6)$ lies on $AC$
$B(-r\cos \theta +5,-r\sin \theta +6)$ lies on $AB$
$\therefore r(\cos \theta +2\sin \theta )=-1$
and $r(2\cos \theta +3\sin \theta )=-1$
Dividing, we get
$\cos \theta +2\sin \theta =2\cos \theta +3\sin \theta$
$\therefore \tan \theta =-1$ and hence the line $BC$ is $x+y=11$