I is true. Rewrite the equation of $L_2$ to the slope-intercept form, we have
$\begin{array}{rcl}
bx + y + c & = & 0 \\
y & = & -bx – c
\end{array}$
From the graph, the $y$-intercept of $L_2$ is positive. Then, we have
$\begin{array}{rcl}
-c & > & 0 \\
c & < & 0
\end{array}$
II is true. Rewrite the equation of $L_1$ to the slope-intercept form, we have
$\begin{array}{rcl}
x + ay + b & = & 0 \\
ay & = & -x -b \\
y & = & \dfrac{-1}{a}x – \dfrac{b}{a}
\end{array}$
From the graph, the slope of $L_1$ is positive. Then, we have
$\begin{array}{rcl}
\dfrac{-1}{a} & > & 0 \\
a & < & 0
\end{array}$
From the graph, the $y$-intercept of $L_1$ is positive. Then, we have
$\begin{array}{rcl}
\dfrac{-b}{a} & > & 0 \\
b & > & 0
\end{array}$
Therefore, we have
$\begin{array}{rcl}
ab & < & 0 \\
ab & < & 1
\end{array}$
III is not true. By comparing the $y$-intercepts of $L_1$ and $L_2$, we have
$\begin{array}{rcl}
-c & > & \dfrac{-b}{a} \\
c & < & \dfrac{b}{a} \\
ac & > & b
\end{array}$
