* A ray of light travels from point $A(-3,5)$ to $l: 3 x-4 y+4=0$, and then reflects to point $B(2,15)$. Find: the length of the path of this ray of light from $A$ to $B$.
Analysis of the line $l: 3 x-4 y+4=0$ shows that its slope is $\frac{3}{4}$. A line $a \perp l$ is drawn through point $A$, and the equation of $a$ is $4 x+3 y-3=0$. Solving the equations of lines $l$ and $a$ gives their intersection point as $C(0, 1)$. If $A^{\prime}$ is symmetric to $A$ with respect to $l$, then $C$ is the midpoint of $A A^{\prime}$. Using the midpoint formula, the coordinates of $A^{\prime}$ are found to be $(3, -3)$. If the reflection point of the light on $l$ is $P$, then $P$ lies on the line $A^{\prime} B$. Therefore, the length of the light path from $A$ to $B$ is
$|A P|+|P B|=\left|\Lambda^{\prime} P\right|+|P B|=\left|\Lambda^{\prime} B\right|=\sqrt{(3-2)^{2}+(-3-15)^{2}}=5 \sqrt{13}$.

13. 2.3 * The coordinates of the three vertices of $\triangle P Q R$ are: $P(-8,5), Q(-15,-19), R(1, -7)$, and the equation of the angle bisector of $\angle P$ can be written as $a x+2 y+c=0, (a, c)=1$. Try to find: $a+c$.
Solution 1: Using the distance formula between two points, we get $|P R|=\sqrt{(-8-1)^{2}+(5+7)^{2}}=15$, $|P Q|=\sqrt{(-8+15)^{2}+(5+19)^{2}}=25$. Let the angle bisector of $\angle P$ intersect $Q R$ at $T$, then by the Angle Bisector Theorem of a triangle, we have $\frac{|P R|}{|P Q|}-\frac{|R T|}{|T Q|}=\frac{15}{25}=\frac{3}{5}$. Using the section formula, we get
$$
x_{T}=\frac{1+\frac{3}{5} \times(-15)}{1+\frac{3}{5}}=-5, y_{T}=\frac{-7+\frac{3}{5} \times(-19)}{1+\frac{3}{5}}=-\frac{23}{2} \text {. }
$$

The equation of line $P T$ is $\frac{y-5}{-\frac{23}{2}-5}=\frac{x+8}{-5+8}$, which simplifies to $11 x+2 y+78=0$. Thus, $a=11, c=78$. Therefore, $a+c=11+78=89$.

Solution 2: Let the slope of the angle bisector of $\angle P$ be $k$, and the slopes of lines $P R$ and $P Q$ be $k_{P R}$ and $k_{P Q}$, respectively, then
$$
k_{P R}=\frac{-7-5}{1+8}=-\frac{4}{3}, k_{P Q}=\frac{-19-5}{-15+8}=\frac{24}{7} .
$$

Using the formula for the angle between two lines and the fact that the angle bisector of $\angle P$ forms equal angles with lines $P R$ and $P Q$, we get $\frac{-\frac{4}{3}-k}{1-\frac{4}{3} k} = \frac{k-\frac{24}{7}}{1+\frac{24}{7} k}$. This simplifies to $22 k^{2}+117 k-22=0$. Solving for $k$, we get $k=-\frac{11}{2}, k=\frac{2}{11}$. Since we are looking for the internal angle bisector of $\triangle P Q R$, we discard $k=\frac{2}{11}$. Thus, the equation of the angle bisector of $\angle P$ is: $y-5=-\frac{11}{2}(x+8)$, which simplifies to $11 x+2 y+78=0$. Therefore, $a=11, c=78, a+c=11+78=89$.