Answer
Lines $L_1$ and $L_2$ are perpendicular.
Work Step by Step
If you let $m_1$ be equivalent to $-\frac{1}{m^2}$, which can be written as $$m_1 = -\frac{1}{m^2}$$
then that would mean the following:
$$\begin{matrix}
m_1 = -\frac{1}{m^2}\\\\
m_1 = \frac{-1}{m^2}\\\\
m_1(m_2) = \frac{-1}{m^2}(m_2)\\\\
m_1m_2 = -1
\end{matrix}$$
Next, if you consider $L_3$ to the line that has $m_3$ as its slope while it is perpendicular to $L_1$, then that would mean the following (with it most likely sounding like $-1 \times 1 = -1$): $$m_1m_3=-1$$
Since $m_1m_3=-1$, then that would mean that $$m_2=m_3$$
but it also means that lines $L_2$ and $L_3$ are parallel.
So, with that being said, lines $L_1$ and $L_2$ are perpendicular.
