smc

$4n=m$, as stated in the question. In the line $L_1$, draw a triangle with the coordinates $(0,0)$, $(1,0)$, and $(1,m)$. Then $m=\tan ({\theta }_1)$. Similarly, $n=\tan ({\theta }_2)$. Since $4n=m$ and ${\theta }_1=2{\theta }_2$, $\tan (2{\theta }_2)=4\tan ({\theta }_2)$. Using the angle addition formula for tangents, $\frac{2\tan ({\theta }_2)}{1-{\tan }^2({\theta }_2)}=4\tan ({\theta }_2)$. Solving, we have $\tan ({\theta }_2)=0,\frac{\sqrt{2}}{2}$. But line $L_1$ is not horizontal, so therefore $(m,n)=(2\sqrt{2},\frac{\sqrt{2}}{2})$. Looking at the answer choices, it seems the answer is $(2\sqrt{2})(\frac{\sqrt{2}}{2})=\boxed{2}$.