If the lines $3x–y+1=0$ and $x–2y+3=0$ are equally inclined to the line $y = mx$, then the value of m is given by 

It $\theta$ is the angle between the lines then $\tan \theta =\left|\frac{m-3}{1+3m}\right|=\left|\frac{m-(1/2)}{1+m(1/2)}\right|$
$\begin{array}{l}\Rightarrow (m-3)(2+m)=\pm (2m-1)(3m+1)\\ \Rightarrow m^2-m-6=\pm \left(6m^2-m-1\right)\end{array}$
$\Rightarrow 7m^2-2m-7=0 or m^2+1=0$