2.1.1. A ray given by the equation $l_{1}: 3 x+4 y-18=0$ falls on the line $l: 3 x+2 y-12=0$, which is capable of reflecting rays. Form the equation of the reflected ray.
Solution. Since the angle of incidence of the ray is equal to the angle of reflection of the ray, then $\angle \phi_{1}=\angle \phi_{2}$, i.e., $\operatorname{tg} \phi_{1}=\operatorname{tg} \phi_{2}$ (Fig. 31).

![](https://cdn.mathpix.com/cropped/2024_05_22_e915d089cc29fd681be5g-120.jpg?height=577&width=635&top_left_y=574&top_left_x=411)

Fig. 31

The equation of the reflected ray - the line $l_{2}$ - is sought in the form:

$$
y-y_{A}=k_{2}\left(x-x_{A}\right)
$$

To find the coordinates of point $A$, we solve the system of equations:

$$
\begin{aligned}
& 3 x+2 y-12=0 \\
& 3 x+4 y-18=0 .
\end{aligned}
$$

Subtracting, we find: $-2 y+6=0, y=3$ and $3 x=12-2 y=12-2 \cdot 3=6$, $x=2$, i.e., $x_{A}=2$ and $y_{A}=3$.

We find the angular coefficients of the lines $l$ and $l_{1}$:

$$
\begin{array}{lll}
l: 2 y=-3 x+12, & y=-\frac{3}{2} x+6, & K=-\frac{3}{2} \\
l_{1}: 4 y=-3 x+18, & y=-\frac{3}{4} x+\frac{9}{2}, & K_{1}=-\frac{3}{4} .
\end{array}
$$

We write the tangent of the angle between the lines $l$ and $l_{1}$:

$$
\operatorname{tg} \phi_{1}=\frac{K_{1}-K}{1+K_{1} K_{2}}=\frac{-\frac{3}{4}--\frac{3}{2}}{1+-\frac{3}{4}-\frac{3}{2}}=\frac{6}{17}
$$

To find the angular coefficient of the line $l_{2}$, we write the tangent of the angle between the lines $l$ and $l_{2}$ and take into account that $\operatorname{tg} \phi_{1}=\operatorname{tg} \phi_{2}$:

$$
\operatorname{tg} \phi_{2}=\frac{K-K_{2}}{1+K K_{2}} \quad \text { or } \quad \frac{-\frac{3}{2}-K_{2}}{1+-\frac{3}{2} K_{2}}=\frac{6}{17}
$$

Then

$$
17-\frac{3}{2}-K_{2}=6 \quad 1-\frac{3}{2} K_{2}
$$

From this, $K_{2}=-\frac{63}{16}$. Then the desired equation of the reflected ray is:

$$
y-3=-\frac{63}{16}(x-2) \text { or } 63 x+16 y-174=0 \text {. }
$$