I is true. According to the figure, the $x$ intercept of the straight line $ax+by+c=0$ is negative. Note that the $x$ intercept of the straight line $ax+by+c=0$ is $\dfrac{-c}{a}$. Hence, we have
$\begin{array}{rcl}
\dfrac{-c}{a} & < & 0 \\
\dfrac{c}{a} & > & 0 \\
\therefore ac & > & 0
\end{array}$
II is false. According to the figure, the $x$ intercept of the straight line $kx+ly+m=0$ is positive. Note that the $x$ intercept of the straight line $mx+ly+m=0$ is $\dfrac{-m}{k}$. Hence, we have
$\begin{array}{rcl}
\dfrac{-m}{k} & > & 0 \\
\dfrac{m}{k} & < & 0 \\
\therefore mk & < & 0
\end{array}$
III is false. If $am=ck$ is true, then we have
$\begin{array}{rcl}
am & = & ck \\
\dfrac{-m}{k} & = & \dfrac{-c}{a}
\end{array}$
That means the two $x$ intercepts of the two straight lines are the same. However, according to the graph, the two $x$ intercepts are two different points.
IV is true. Note that the $y$ intercept of the straight line $ax+by+c=0$ is $\dfrac{-c}{b}$ and that of the straight line $kx+ly+m=0$ is $\dfrac{-m}{l}$. Since they intersect at a point on the $y$-axis, the two $y$ intercepts are the same. Hence, we have
$\begin{array}{rcl}
\dfrac{-c}{b} & = & \dfrac{-m}{l} \\
bm & = & cl
\end{array}$
