13.2.11 ** Let $\triangle \Lambda B C$ be a right-angled triangle in the $x y$ plane, with $C$ being the right angle, and the hypotenuse $A B$ having a length of 60. The medians drawn from $A$ and $B$ lie on the lines $y=x+3, y=2 x+4$, respectively. Find $S_{\triangle A B C}$.
As shown in the figure, let the median $A D=3 t, B E=3 s$, and the centroid be $G$. The acute angle between $A D$ and $B E$ is $\angle A G E=\theta$, and let $A C=2 p, B C=2 q$. Since the equations of the lines $A D$ and $B E$ are $y=x+3$ and $y=2x+4$, the tangent of the acute angle between these two lines is $\tan \theta=\left|\frac{1-2}{1+1 \times 2}\right|=\frac{1}{3}$. Thus, $\sin \theta=\frac{1}{\sqrt{10}}$. Let the area of $\triangle A B C$ be $S_{\triangle A B C}=S$. Then $S=6 S_{\triangle A G E}$. Also, $S_{\triangle A G E}=\frac{1}{2} A G \cdot G E \cdot \sin \theta=\frac{1}{2} \cdot 2 t \cdot s \cdot \frac{1}{\sqrt{10}}=\frac{s t}{\sqrt{10}}$. Therefore, $S=\frac{1}{2} A C \cdot B C=2 p q$. By the Pythagorean theorem, $A C^{2}+B C^{2}=4 p^{2}+4 q^{2}=A B^{2}=3600$, so $p^{2}+q^{2}=900$. Also, $A C^{2}+C D^{2}=A D^{2}, B C^{2}+C E^{2}=B E^{2}$, i.e., 
$$
\left\{\begin{array}{l}
4 p^{2}+q^{2}=9 t^{2} \\
4 q^{2}+p^{2}=9 s^{2}
\end{array}\right.
$$
Thus, from $S=6 S_{\triangle A G E}=\frac{6 s t}{\sqrt{10}}$, we get
$$
\begin{aligned}
S^{2} & =\frac{36}{10} s^{2} t^{2}=\frac{18}{5} \cdot \frac{4 p^{2}+q^{2}}{9} \cdot \frac{4 q^{2}+p^{2}}{9}=\frac{2}{45}\left[\left(2 p^{2}+2 q^{2}\right)^{2}+9 p^{2} q^{2}\right] \\
& =\frac{2}{45}\left(1800^{2}+\frac{9}{4} S^{2}\right)=144000+\frac{1}{10} S^{2} \text {. Solving for } S \text{, we get } S=400 .
\end{aligned}
$$