2008-II-31

$\begin{array}{cl}
= & \tan 30^\circ \\
= & \dfrac{1}{\sqrt{3}}
\end{array}$

由於 $L_1 \perp L_2$，所以 $L_2$ 的斜率

$\begin{array}{cl}
= & -1 \div \dfrac{1}{\sqrt{3}} \\
= & -\sqrt{3}
\end{array}$

所以，$L_2$ 的方程為

$\begin{array}{rcl}
y – 0 & = & -\sqrt{3}(x-0) \\
y & = & -\sqrt{3} x \\
\sqrt{3} x + y & = & 0
\end{array}$