4. As shown in Figure 5, the quadrilateral $ABCD$ is a schematic diagram of a desert area, with points $A$ and $B$ on the $x$-axis, $E(2,6)$, $F(3,4)$. The broken line $OFE$ is a water channel flowing through this desert. The desert to the east of the water channel is contracted for greening by person A, and the desert to the west of the water channel is contracted for greening by person B.
Now, A and B have agreed: in
the greening plan, the water
channel flowing through the
desert needs to be straightened,
and the areas of the desert contracted
by A and B must remain unchanged.

If a point $P$ is to be found on $AB$ such that the water channel is straightened to $EP$, then the coordinates of point $P$ are $\qquad$
4. $\left(\frac{5}{3}, 0\right)$.

As shown in Figure 14, connect $O E$, draw $F P / / O E$ intersecting $A B$ at point $P$, connect $E P$ intersecting $O F$ at point $G$.
Since $O E / / P F$, then
$S_{\triangle O E F}=S_{\triangle O B P}$.
Thus, $S_{\triangle O E P}-S_{\triangle O E C}$
$$
=S_{\triangle O B P}-S_{\triangle O E C} \text {, }
$$

which means $S_{\triangle E F G}=S_{\triangle O C P}$.
Therefore, $E P$ is the straight line, and point $P$ is the required point.
It is easy to find that the equation of line $O E$ is $y=3 x$.
Since $O E / / P F$, the equation of line $P F$ can be set as $y=3 x+b$.
Given $F(3,4)$, we have $4=3 \times 3+b$, which means $b=-5$. Therefore, the equation of line $P F$ is $y=3 x-5$.
When $y=0$, $3 x-5=0, x=\frac{5}{3}$.
Thus, the coordinates of point $P$ are $\left(\frac{5}{3}, 0\right)$.