Example 6 Light passes through point $A(2,3)$, hits the line $l: x+y+1=0$, and after reflection passes through point $B(1,1)$, find the equations of the incident and reflected rays.
Solution 1: Since the angle of incidence equals the angle of reflection, the lines of incidence and reflection are symmetrical about line $l$. To find the equation of the reflection line $A^{\prime} B$, we only need to find the symmetric point $A^{\prime}$ of point $A$ with respect to line $l$.
Let the symmetric point of $A(2,3)$ about $l: x+y+1=0$ be $A^{\prime}\left(x^{\prime}, y^{\prime}\right)$,
then $\left\{\begin{array}{l}\frac{y^{\prime}-3}{x^{\prime}-2}=1, \\ \frac{x^{\prime}+2}{2}+\frac{y^{\prime}+3}{2}+1=0 .\end{array}\right.$ Solving this, we get $\left\{\begin{array}{l}x=-4, \\ y^{\prime}=-3 .\end{array}\right.$
Therefore, the equation of the reflection line $A^{\prime} B$ is $4 x-5 y+1=0$.
Since the incident line is not parallel to the $y$-axis, let the slope of the incident line be $k$. By the principle that the angle of incidence equals the angle of reflection, we have:
$\frac{k+1}{1-k}=\frac{-1-\frac{4}{5}}{1-\frac{4}{5}}$, which simplifies to $k=\frac{5}{4}$.
Thus, the equation of the incident line is $5 x-4 y+2=0$.
Solution 2: Using the property that the symmetric point of point $A(x, y)$ about the line $x \pm y+c=0$ is $A^{\prime}(\mp y-c, \mp(x+c))$ to solve.

Since the incident line is not parallel to the $y$-axis, let the equation of the incident line be $y-3=k(x-2)$.

Since the reflection line and the line of reflection are symmetrical about $l$, from $x+y+1=0$, we get $x=-y-1, y=-x-1$, thus proving that the equation of the reflection line is $x-k y+4-3 k=0$.
Since the reflection line passes through $B(1,1)$, then $k=\frac{5}{4}$.
Therefore, the equations of the incident line and the reflection line are $5 x-4 y+2=0, 4 x-5 y+1=0$, respectively.