## Task B-3.1.

Given is a triangle $A B C$. The side $\overline{B C}$ of this triangle lies on the line $2 x+y+10=0$, the side $\overline{A C}$ lies on the line $2 x-y-2=0$, and the median from vertex $A$ lies on the line $2 x-7 y+10=0$. Determine the coordinates of the vertices of triangle $A B C$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b05efd680c81b2b8cbfg-14.jpg?height=605&width=1033&top_left_y=871&top_left_x=520)

Let $a$ be the line $BC$, $b$ be the line $AC$, and $t_{a}$ be the line $AP$ on which the median from vertex $A$ lies. Then $a \cap b=\{C\}$, and $b \cap t_{a}=\{A\}$.

Thus, the coordinates of vertex $C$ will be obtained as the solution to the system

$$
\begin{gathered}
2 x+y+10=0 \\
2 x-y-2=0
\end{gathered}
$$

By adding the equations, we get $x=-2$, and then from one of the equations, $y=-6$. Therefore, vertex $C$ is $C(-2,-6)$.

The coordinates of vertex $A$ will be obtained as the solution to the system

$$
\begin{gathered}
2 x-7 y+10=0 \\
2 x-y-2=0
\end{gathered}
$$

By subtracting the equations of the system, we get $y=2$, and then from one of the equations, $x=2$. Therefore, vertex $A$ is $A(2,2)$.

The midpoint of side $\overline{BC}$ is the point $P\left(x_{P}, y_{P}\right)$, and for its coordinates, we have

$$
x_{P}=\frac{x_{B}+x_{C}}{2}=\frac{x_{B}-2}{2}
$$

$$
y_{P}=\frac{y_{B}+y_{C}}{2}=\frac{y_{B}-6}{2}
$$

Since point $P$ lies on the median, i.e., on the line $t_{a}$, and point $B$ lies on the line $a$, their coordinates satisfy the equations of these lines. Therefore, we have:

$$
\begin{gathered}
2 \cdot \frac{x_{B}-2}{2}-7 \cdot \frac{y_{B}-6}{2}+10=0 \\
2 x_{B}+y_{B}+10=0
\end{gathered}
$$

After simplifying, we get the system of equations

$$
\begin{gathered}
2 x_{B}-7 y_{B}+58=0 \\
2 x_{B}+y_{B}+10=0
\end{gathered}
$$

By subtracting the equations of this system, we get $y_{B}=6$, and then from one of the equations, $x_{B}=-8$. Therefore, vertex $B$ is $B(-8,6)$.