 Coordinate Method ] Complex [ The ratio of areas of triangles with a common base or common height] Class

In quadrilateral $A B C D$, find a point $E$ such that the ratio of the areas of triangles $E A B$ and $E C D$ is $1: 2$, and the ratio of the areas of triangles $E A D$ and $E B C$ is $3: 4$, given the coordinates of all its vertices: $A(-2 ;-4)$, $B(-2 ; 3)$, $C(4 ; 6)$, $D(4 ;-1)$.
Since

$$
\overrightarrow{A B}=\overrightarrow{(-2-(-2): 3-(-4))}=\overrightarrow{(0: r)} \overrightarrow{D C}=\overrightarrow{(+-4: 6-(-1))}=\overrightarrow{(0: r)}=\overrightarrow{A B}
$$

then the quadrilateral $A B C D$ is a parallelogram, with sides $A B$ and $C D$ parallel to the y-axis. Let $E$ be the desired point. The triangles $E A B$ and $E C D$ have equal and parallel bases $A B$ and $C D$, so the equality $\frac{S_{\triangle E, A B}}{S_{\triangle E C D}}=\frac{1}{2}$ is equivalent to the fact that their common vertex $E$ lies on a line parallel to the bases and at a distance from them such that the ratio of these distances is $\frac{1}{2}$, i.e., on the y-axis.

Let the line $B C$ intersect the y-axis at point $P$, and the line $A D$ at point $Q$. If $O$ is the origin, then $O P=3$ and $O Q=4$.

The triangles $E A D$ and $E B C$ have equal and parallel bases $A D$ and $B C$, so the equality $\frac{S_{\triangle E, A D}}{S_{\triangle E B C}}=\frac{3}{4}$
is equivalent to the fact that their common vertex $E$ lies on a line parallel to the bases and at a distance from them such that the ratio of these distances is $\frac{3}{4}=\frac{O P}{O Q}$, i.e., on the line passing through point $O$.

Therefore, the point $E$ is the origin.

## Answer

$E(0 ; 0)$