$25 \cdot 1$ On the plane, there are points $A(-2,1), B(1,4), C(4,-3)$, point $D$ is on line $A B$, $\frac{A D}{B D}=\frac{1}{2}$, extend $D C$ to $E$, such that $\frac{|C E|}{|E D|}=\frac{1}{4}$. Then the coordinates of point $E$ are (A) $(0,1)$.
(B) $(0,1)$ or $\left(\frac{6}{3}, \frac{11}{3}\right)$.
(C) $\left(-\frac{8}{3}, \frac{11}{3}\right)$.
(D) $\left(\frac{17}{3},-\frac{14}{3}\right)$.
(E) $\left(-\frac{16}{5},-\frac{11}{5}\right)$.
[Solution] As long as a fairly accurate sketch is drawn based on the given data, it can be verified or determined that the coordinates of point $E$ are $\left(\frac{17}{3},-\frac{14}{3}\right)$.
Therefore, the answer is $(D)$.