[Properties and characteristics of a parallelogram]  [Vectors help solve tasks]

Given points $A(-6; -1), B(1; 2)$, and $C(-3; -2)$. Find the coordinates of vertex $M$ of parallelogram $A B M C$.
## First method.

The coordinates of the midpoint $K\left(x_{0} ; y_{0}\right)$ of the diagonal $B C$ of the parallelogram $A B M C$ are the arithmetic means of the corresponding coordinates of the endpoints of the segment $B C$, i.e.,

$$
x_{0}=\frac{1-3}{2}=-1, y_{0}=\frac{2-2}{2}=0
$$

Since the diagonals of a parallelogram bisect each other, $K\left(x_{0} ; y_{0}\right)$ is the midpoint of the segment with endpoints at $A(-6 ;-1)$ and $M\left(x_{1} ; y_{1}\right)$. Therefore,

$$
x_{0}=\frac{-6+x_{1}}{2}=-1, y_{0}=\frac{-1+y_{1}}{2}=0
$$

From this, we find that $x_{1}=4, y_{1}=1$.

Second method.

Let $x_{1}, y_{1}$ be the coordinates of point $M$. If $A B M C$ is a parallelogram, then $\overrightarrow{B M}=\overrightarrow{A C}$, and since

$$
\overrightarrow{B M}=\overrightarrow{\left(x_{1}-1 ; y_{1}-2\right)}, \overrightarrow{A C}=\overline{(-3-(-6) ;-2-(-1)}=\overrightarrow{(3 ; 1)}
$$

From this, we find that $x_{1}=4, y_{1}=1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_243bf411927c48073e50g-26.jpg?height=509&width=806&top_left_y=123&top_left_x=635)

## Answer

$M(4 ; 1)$.