Example 4 Two parallel lines $l_{1}, l_{2}$ are at a distance of $d$, they pass through points $M(-2,-2), N(1,3)$ respectively, and rotate around $M, N$ while remaining parallel. Find the equations of the two lines $l_{1}, l_{2}$ when $d$ is at its maximum value.
Let the equations of the lines $l_{1}, l_{2}$ be: $y=k x+b_{1}, y=k x+b_{2}$.
Since $l_{1}$ passes through $M(-2,-2)$ and $l_{2}$ passes through $N(1,3)$, we have:
$$
\left\{\begin{array}{l}
-2=-2 k+b_{1}, \\
3=k+b_{2} .
\end{array}\right.
$$

From (5) - (6) we get $b_{1}-b_{2}=3 k-5$, and
$$
d=\frac{\left|b_{1}-b_{2}\right|}{\sqrt{1+k^{2}}}=\frac{|3 k+5|}{\sqrt{1+k^{2}}} .
$$

Rearranging gives $\left(d^{2}-9\right) k^{2}+30 k+d^{2}-25=0$.
Since $k \in \mathbb{R}$, we have $\triangle=30^{2}-4\left(d^{2}-9\right)\left(d^{2}-25\right) \geqslant 0$, and since $d>0$, solving gives $0<d \leqslant \sqrt{34}$.

When $l_{1}, l_{2}$ are parallel to the $y$-axis, $d=3<\sqrt{34}$, so when $d=\sqrt{34}$, equation (7) becomes $25 k^{2}+30 k+9=0$, solving which gives $k=-\frac{3}{5}$. Substituting into equations (5) and (6), we get $b_{1}=-\frac{16}{5}$, $b_{2}=\frac{18}{5}$.
Therefore, $l_{1}: 3 x+5 y+16=0, l_{2}: 3 x+5 y-18=0$.