8,9

Given points $A(-2 ; 1), B(2 ; 5)$, and $C(4 ;-1)$. Point $D$ lies on the extension of median $A M$ beyond point $M$, and quadrilateral $A B D C$ is a parallelogram. Find the coordinates of point $D$.
## First method.

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

$$
x_{0}=\frac{2+4}{2}=3, y_{0}=\frac{5-1}{2}=2
$$

Since the diagonals of a parallelogram are bisected by the point of intersection, $M\left(x_{0} ; y_{0}\right)$ is the midpoint of the segment with endpoints at points $A(-2 ; 1)$ and $D\left(x_{1} ; y_{1}\right)$. Therefore,

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

From this, we find that $x_{1}=8, y_{1}=3$.

## Second method.

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

$$
\overrightarrow{B D}=\overrightarrow{\left(x_{1}-2 ; y_{1}-5\right)}, \overrightarrow{A C}=\overrightarrow{(4-(-2) ;-1-1)}=\overrightarrow{(6 ;-2)}
$$

then

$$
x_{1}-2=6, y_{1}-5=-2 \text {. }
$$

From this, we find that $x_{1}=8, y_{1}=3$.

## Answer

$(8 ; 3)$.