2. In the right triangle $ABC$, the hypotenuse $AB=60$, and the medians of sides $CA$, $CB$ lie on the lines with equations $y=x+3$, $y=2x+4$, respectively. Then the area of $\triangle ABC$ is $\qquad$
2.400

Analysis: Let the centroid of the triangle be $G$, and the midpoint of $A B$ be $R$. Connect $G R, G B, G A$. It is easy to know that $G R=10$. Let $\angle A G B=\alpha$, then $\sin \alpha=\frac{1}{\sqrt{10}}, \cos \alpha=-\frac{3}{\sqrt{10}}$. Let $G A=x, G B=y$. By the cosine rule, we have $x^{2}+y^{2}-2 x y \cos \alpha=A B^{2}$. By the parallelogram law, we have $2\left(x^{2}+y^{2}\right)=A B^{2}+4 R G^{2}$. Thus, $2 x y \cos \alpha=-1600$, so $S_{\triangle} A B G=\frac{1}{2} x y \sin \alpha=\frac{400}{3}$. Therefore, $S_{\triangle} A B C=3 S_{\triangle} A B G=400$.