78. As shown in the figure, in the Cartesian coordinate system, $\mathrm{O}(0,0), \mathrm{A}(0,5), \mathrm{B}(6,5), \mathrm{D}(6,0)$, a line 1 passing through point $\mathrm{M}(5,6)$ divides the rectangle $O A B C$ into two parts with an area ratio of $2: 3$. Find the equation of line 1.
Solution: Let line 1 intersect segment $\mathrm{AB}$ at point $\mathrm{D}$, and intersect the $\mathrm{x}$-axis at point $\mathrm{E}$.
From the diagram, when point $\mathrm{E}$ is to the left of point $\mathrm{O}$ (including point $\mathrm{O}$), the rectangle is divided into a triangle and a pentagon, and when point $\mathrm{E}$ coincides with $\mathrm{O}$, the area of the triangle is maximized. At this time, the equation of line 1 is
$$
\begin{array}{c}
y=\frac{5}{6} x, \\
\left(\frac{25}{6}, 5\right), \\
\frac{1}{2} \times 5 \times \frac{25}{6}=\frac{125}{12}<\frac{2}{3+2} \times 5 \times 6,
\end{array}
$$

Let $y=5$, we can get the coordinates of point $D$ as

The area of the triangle is
Therefore, point $\mathrm{E}$ is to the right of $\mathrm{O}$.
Similarly, point $\mathrm{E}$ is to the left of $\mathrm{C}$.
Thus, point $\mathrm{E}$ lies on segment $\mathrm{OC}$. At this time, line 1 divides the rectangle into two trapezoids. Let $\mathrm{E}(\mathrm{a}, 0)$, then
the equation of line 1 is

Let $y=5$, we can get the coordinates of point $\mathrm{D}$ as

Therefore, the ratio of the areas of the two trapezoids is

When $\frac{\frac{25+a}{6}+a}{6-a+\left(6-\frac{25+a}{6}\right)}=\frac{2}{3}$, solving gives $a=\frac{19}{35}$,
at this time, the equation of line 1 is
$$
y=\frac{35}{26} x-\frac{19}{26} \text {. }
$$

When $\frac{\frac{25+a}{6}+a}{6-a+\left(6-\frac{25+a}{6}\right)}=\frac{3}{2}$, solving gives
at this time, the equation of line 1 is
$$
\begin{array}{c}
y=\frac{6}{5-a}(x-a), \\
\left(\frac{25+a}{6}, 5\right), \\
\frac{25+a}{6}+a \\
6-a+\left(6-\frac{25+a}{6}\right)
\end{array}
$$