$25 \cdot 29$ In the $xy$ plane, the horizontal and vertical lines connecting the points $(0,0)$, $(0,3)$, $(3,3)$, $(3,1)$, $(5,1)$, and $(5,0)$ form an "L"-shaped region (see figure). The slope of the line passing through the origin and dividing the area of this figure in half is
(A) $\frac{2}{7}$.
(B) $\frac{1}{3}$.
(C) $\frac{2}{3}$.
(D) $\frac{3}{4}$.
(E) $\frac{7}{9}$.
［Solution］From the given conditions, it is easy to find that
$$
S_{A B C D D O E}=11 \text {, }
$$

Also, $S_{\text {OBCD }}=\frac{7}{2}<\frac{1}{2} S_{A B C D O E}$,
we know that the area bisector line $O P$ intersects $A B$ at point $F$ above point $B$. Let $A F=t$,
$\because \quad S_{\text {trapezoid }}$.
$\therefore \quad$ The coordinates of point $F$ are $\left(3, \frac{7}{3}\right)$, so $k_{O P}=\frac{7}{9}$. Therefore, the answer is $(E)$.