[Coordinate method on the plane]

Find the coordinates of the vertices of a triangle, the sides of which lie on the lines $2 x+y-6=0, x-y+4=0$ and $y+1$ $=0$.
Solving the system of equations

$$
\left\{\begin{array}{l}
2 x+y-6=0 \\
x-y+4=0
\end{array}\right.
$$

we find the coordinates of the point $A\left(x_{1} ; y_{1}\right)$ of intersection of the given lines: $x_{1}=\frac{2}{3}, y_{1}=\frac{14}{3}$.

Similarly, we will find the other vertices of the triangle.

## Answer

$\left(\frac{2}{3} ; \frac{14}{3}\right),(-5 ;-1),\left(\frac{7}{2} ;-1\right)$

## Problem