Let [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) $ABC$ be a [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle) in the xy-plane with a right angle at $C_{}$. Given that the length of the [hypotenuse](https://artofproblemsolving.com/wiki/index.php/Hypotenuse) $AB$ is $60$, and that the [medians](https://artofproblemsolving.com/wiki/index.php/Median) through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$.
Let $\theta_1$ be the angle that the median through $A$ makes with the positive $x$-axis, and let $\theta_2$ be the angle that the median through $B$ makes with the positive $x$-axis.  The tangents of these two angles are the slopes of the respective medians; in other words, $\tan \theta_1 = 1$, and $\tan \theta_2 =2$.
Let $\theta$ be the angle between the medians through $A$ and $B$.  Then by the [ tangent angle subtraction formula](https://artofproblemsolving.com/wiki/index.php/Trigonometric_identities#Angle_addition_identities),
\[\tan \theta = \tan (\theta_2 - \theta_1) = \frac{\tan \theta_2 - \tan \theta_1}{1 + \tan \theta_1 \tan \theta_2} = \frac{2-1}{1 + 2 \cdot 1 } = \frac{1}{3}.\]


Let $M$ be the midpoint of $BC,$ let $N$ be the midpoint of $AC,$ and let $G$ be the intersection of these two medians.  Let $BC = a,$ and let $AC = b$; let $\alpha = \angle MAC$, and let $\beta = \angle BNC$.
With this setup, $\theta = \beta - \alpha$.  The reason is that $\beta$ is an exterior angle of $\triangle AGN$, and thus $\beta = \alpha + \angle AGN = \alpha + \theta$.
We also know that $\tan \alpha = \dfrac{(1/2) a}{b} = \dfrac{a}{2b}$, and $\tan \beta = \dfrac{a}{(1/2)b} = \dfrac{2a}{b}$.  Then
\begin{align*} \frac{1}{3} &= \tan \theta \\ &= \tan (\beta - \alpha) \\ &= \frac{\tan \beta - \tan \alpha}{1 + \tan \beta \tan \alpha} \\ &= \frac{2 \frac{a}{b} - \frac{1}{2} \cdot \frac{a}{b}}{1 + (a/b)^2} \\ &= \frac{(3/2) \cdot (a/b)}{1 + (a/b)^2} . \end{align*}
Multiplying numerator and denominator by $b^2,$ we learn that
\[\frac{1}{3} = \frac{(3/2) ab}{a^2 + b^2} = \frac{(3/2) ab}{60^2},\]
since the hypotenuse of $\triangle ABC$ has length 60.  Solving for $\frac{1}{2}ab$, we find that
\[\frac{1}{2} ab = \frac{1}{9} \cdot 60^2 = (60/3)^2 = 20^2 = 400.\]