Ans: D
Note that the slope of $L_1$
Note that the slope of $L_1$
$\begin{array}{cl}
= & \tan 30^\circ \\
= & \dfrac{1}{\sqrt{3}}
\end{array}$
Since $L_1 \perp L_2$, the slope of $L_2$
$\begin{array}{cl}
= & -1 \div \dfrac{1}{\sqrt{3}} \\
= & -\sqrt{3}
\end{array}$
Therefore, the equation of $L_2$ is
$\begin{array}{rcl}
y – 0 & = & -\sqrt{3}(x-0) \\
y & = & -\sqrt{3} x \\
\sqrt{3} x + y & = & 0
\end{array}$
