## Task B-4.6.

The equations of the lines on which two sides of the triangle lie are $AB: 3x + y - 3 = 0$, $AC$: $3x + 4y = 0$. If the equation of the angle bisector of $\beta$ is $s_{\beta}: x - y + 5 = 0$, determine the coordinates of the vertices of the triangle $ABC$.
## Solution.

$\operatorname{Vertex} A$ is the intersection of lines $A B$ and $A C, A=\left(\frac{4}{3},-1\right)$.

Vertex $B$ is the intersection of lines $A B$ and $s_{\beta}, B=\left(-\frac{1}{2}, \frac{9}{2}\right)$.

Let the slopes of lines $A B, s_{\beta}, B C$ be denoted by $k_{A B}, k_{s_{\beta}}, k_{B C}$. Due to the property of the angle bisector, the angles $\angle\left(B A, s_{\beta}\right)$ and $\angle\left(s_{\beta}, B C\right)$ are equal, and thus their tangents are also equal, so we have:

$$
\begin{aligned}
\left|\frac{k_{A B}-k_{s_{\beta}}}{1+k_{A B} \cdot k_{s_{\beta}}}\right| & =\left|\frac{k_{s_{\beta}}-k_{B C}}{1+k_{B C} \cdot k_{s_{\beta}}}\right| \\
\left|\frac{-3-1}{1+(-3) \cdot 1}\right| & =\left|\frac{1-k_{B C}}{1+1 \cdot k_{B C}}\right|
\end{aligned}
$$

This equation has two solutions:

$$
k_{B C}=-\frac{1}{3}, \quad k_{B C}=-3
$$

but only $k_{B C}=-\frac{1}{3}$ is the correct solution because in the other case, the lines on which two sides of the triangle lie have the same slope, which is impossible.

(1 point)

## Note:

By correctly ordering the slopes, the absolute value in ( $\dagger$ ) can be omitted, and only one valid solution is immediately obtained. In this case, there should be a sketch from which the order is visible, and thus all 3 points can be given for determining $k_{B C}$.

The equation of line $B C$ through point $B$ with the known slope $k_{B C}=-\frac{1}{3}$ is $B C: y-\frac{9}{2}=-\frac{1}{3}\left(x+\frac{1}{2}\right)$, or $B C: x+3 y-13=0$.

(2 points)

Finally, vertex $C$ is the intersection of lines $B C$ and $A C, C=\left(-\frac{52}{5}, \frac{39}{5}\right)$.