Example 2. Find the equation of the line that is parallel to the line $l: 4 x-3 y+8=0$ and forms a triangle with the coordinate axes with an area of 6 square units.

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The family of lines parallel to $l$ has the equation
$$
4 x-3 y+k=0 \text {, }
$$

which can be rewritten as $\frac{x}{-\frac{k}{4}}+\frac{y}{\frac{k}{3}}=1$.
The intercepts of this family of lines on the two axes are
$$
a=-\frac{k}{4}, \quad b=\frac{k}{3} .
$$

According to the problem, we have
$$
\begin{array}{l}
\frac{1}{2}|a b|=6 \\
\left|\left(-\frac{k}{4}\right) \cdot \frac{k}{3}\right|=12, \quad k= \pm 12
\end{array}
$$

The required equations of the lines are
$$
4 x-3 y \pm 12=0
$$