jmc

Translate so the medians are $y=x$, and $y=2x$, then model the points $A:(a,a)$ and $B:(b,2b)$. $(0,0)$ is the centroid, and is the average of the vertices, so $C:(-a-b,-a-2b)$ $AB=60$ so $3600=(a-b{)}^2+(2b-a{)}^2$ $3600=2a^2+5b^2-6ab(1)$ $AC$ and $BC$ are perpendicular, so the product of their slopes is $-1$, giving $\left(\frac{2a+2b}{2a+b}\right)\left(\frac{a+4b}{a+2b}\right)=-1$ $2a^2+5b^2=-\frac{15}{2}ab(2)$ Combining $(1)$ and $(2)$, we get $ab=-\frac{800}{3}$ Using the determinant product for area of a triangle (this simplifies nicely, add columns 1 and 2, add rows 2 and 3), the area is $\left|\frac{3}{2}ab\right|$, so we get the answer to be $\boxed{400}$.