The mirror’s normal is along $(1,1)$, so reflecting the unit direction $u=(\cos30^\circ,\sin30^\circ)=\left(\dfrac{\sqrt3}{2},\dfrac12\right)$ about the line gives $u'=u-2(u\cdot \hat n)\hat n=\left(-\dfrac12,-\dfrac{\sqrt3}{2}\right)$, 

i.e. slope $m'=\sqrt3$. 

The reflected ray through $P$ is $y-y_0=\sqrt3(x-x_0)$. 

Intersecting $y=0$ gives $x=x_0-\dfrac{y_0}{\sqrt3} $

$=\dfrac{\sqrt3}{\sqrt3+1}-\dfrac{1}{\sqrt3(\sqrt3+1)}$

$=\dfrac{2}{3+\sqrt3}$