Question

Find the angles of a triangle whose sides are $x+2y-8=0,3x+y-1=0$ and $x-3y+7=0$

Answer 1

$x+2y-8=0$ $M_1=-1/2$

$3x+y-1=0$ $M_2=-3$

$x+3y+7=0$ $M_3=1/3$

$M_2{M}_3=-1$

$\therefore {90}^o$

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$=\left|\frac{-1/2+3}{1+3/2}\right|$
$=\left|\frac{-1+6}{-5}\right|$

$\Rightarrow \theta ={\tan }^{-1}$

$={45}^o$
$\tan \varphi -\left|\frac{m_1-m_3}{1+m_1{m}_3}\right|=\left|\frac{-1/3-1/3}{1-1/6}\right|=1$

$\varphi =\frac{\pi }{4}$