15. Let $\triangle A B C$ be a right triangle on the $x y$ plane, with $C$ as the right angle, and the chord $A B$ has a length of 60. The medians drawn from $A$ and $B$ lie on the lines $y=x+3$ and $y=2 x+4$, respectively. Find the area of $\triangle A B C$.
15. In this problem, $r=\frac{60}{2}=30$.

From figure (1), it is easy to calculate $\operatorname{tg} \theta=\frac{1}{3}, \sin \theta=\frac{1}{\sqrt{10}}$. As shown in figure (2), taking the midpoints of the three sides as $M, N, O$, with $D$ as the incenter, $p, q, s, t$ are defined as shown in the figure, $\angle \theta=\angle A D M$.
$\operatorname{tg} \theta=\frac{1}{3}, \sin \theta=\frac{1}{\sqrt{10}}$ have been calculated as above.
Since $S_{\triangle A B C}=\delta S_{\triangle A D M}$

Therefore, $S_{\triangle A B C}=6 s t \sin \theta=\frac{6}{10} s t$.
Also, since $\angle C=90^{\circ}$,
(2)
By the Pythagorean theorem, we have
$$
\begin{array}{l}
4 p^{2}+4 Q^{2}=A B^{2}=3600, \\
p^{2}+4 q^{2}=9 s^{2}, \\
4 p^{2}+q^{2}=9 t^{2} .
\end{array}
$$

From (1) and (2), we get
$$
\begin{array}{l}
\left(S_{\triangle A B C}\right)^{2}=\frac{18}{5} s^{2} t^{2} \\
\quad=\frac{18}{5} \cdot \frac{p^{2}+4 q^{2}}{9} \cdot \frac{4 q^{2}+q^{2}}{9} \\
\quad=\frac{2}{45}\left(\left(2 p^{2}+2 q^{2}\right)^{2}+9 p^{2} q^{2}\right) \\
\quad=\frac{2}{45}\left(\left(\frac{3600}{2}\right)^{2}+\frac{9}{4}\left(S_{\triangle A B C}\right)^{2}\right) \\
\quad=144000+\frac{1}{10}\left(S_{\triangle A B C}^{2} .\right. \\
\therefore \frac{9}{10}\left(S_{\triangle A B C}=2=144000 .\right.
\end{array}
$$
$\therefore$ The area of $\triangle A B C$ is 400.
(Edited by Tang Shouwen and Zhi Chunlu, Beijing Mathematical Society)