Computing area of triangle via equations of medians For a triangle $ABC$,
$B=90^\circ , AC=6$, equations of medians through $A$ is $y=2x+4$ and through $C$ is $y=x+3$. What is the area of triangle $ABC$?
I'm really bad at geometry, and to make matters worse, the equations of the medians are given. I have no idea where to start. A hint would be appreciated.


Let denote the sides $|BC|=a$, $|AC|=b$, $|AB|=c$
and medians $|AA_m|=m_a$, $|BB_m|=m_b$, $|CC_m|=m_c$
of $\triangle ABC$.
Also, let 
$x=|GA_m|=\tfrac13|AA_m|$,
$y=|GC_m|=\tfrac13|CC_m|$.
Since $\angle ABC=90^\circ$,
\begin{align}
|BB_m|=m_b=\tfrac12|AC|=\tfrac12b=3
.
\end{align}
It is known that for such right triangle
\begin{align}
a^2+c^2&=b^2
\tag{1}\label{1}
,\\
m_a^2+m_c^2&=5m_b^2
,\\
x^2+y^2&=\tfrac59\,m_b^2
=\tfrac5{36}\,b^2
\tag{2}\label{2}
.
\end{align}
\begin{align}
\triangle GCA_m:\quad
|CA_m|^2&=
|GA_m|^2+|GC|^2
-2|GA_m|\cdot|GC|\cos\theta
,\\
\tfrac14a^2&=
x^2+4y^2-4xy\cos\theta
,\\ 
\tfrac14a^2&=
\tfrac59b^2-3x^2-4xy\cos\theta
\tag{3}\label{3}
.
\end{align}
Similarly,
\begin{align}
\triangle GC_mA:\quad
|AC_m|^2&=
|GC_m|^2+|GA|^2
-2|GC_m|\cdot|GA|\cos\theta
,\\
\tfrac14c^2&=
4x^2+y^2-4xy\cos\theta
,\\ 
\tfrac14c^2&=
\tfrac59b^2-3y^2-4xy\cos\theta
\tag{4}\label{4}
.
\end{align}
Combination of \eqref{3}+\eqref{4} 
with \eqref{1},\eqref{2}
results in
\begin{align}
y&=\frac{b^2}{18\,x\cos\theta}
\tag{5}\label{5}
\end{align}
Substitution of \eqref{5} into \eqref{2}
provides a quadratic equation in $x^2$
with the pair of roots,
which represents two combinations of $x^2,y^2$. 
\begin{align}
x^2&=\tfrac5{72}\,b^2\left(1 \pm \sqrt{1-\frac 1{(\tfrac54\,\cos\theta)^2}}\right)
,\\
y^2&=\tfrac5{72}\,b^2\left(1 \mp \sqrt{1-\frac 1{(\tfrac54\,\cos\theta)^2}}\right)
.
\end{align}
The area of a general $\triangle ABC$
in terms of its medians is
\begin{align}
S&=\frac13\sqrt{4m_a^2m_c^2-(m_a^2+m_c^2-m_b^2)^2}
,
\end{align}
and in case of this right triangle we have
\begin{align}
S&=\frac13\sqrt{4m_a^2m_c^2-(4m_b^2)^2}
.
\end{align}
Using $m_a^2=9x^2$, $m_c^2=9y^2$,
we arrive at a nice 
expression for the area of the right triangle
in terms of its hypotenuse $b$ 
and the angle $\theta$
between its medians $AA_m$ and $CC_m$
\begin{align}
S&=\tfrac13\,b^2\tan\theta
.
\end{align}
The length of hypotenuse is given,
and the angle $\theta=\angle A_mGC=\angle AGC_m$
and its tangent and cosine
can be derived 
from the equations $y=2x+4$, $y=x+3$
of the lines through medians,
\begin{align}
\tan\theta&=\tan(\arctan(2)-\arctan(1))
=\frac{\tan(\arctan(2))-\tan(\arctan(1))}{1+\tan(\arctan(2))\tan(\arctan(1))}
=\frac13
,\\
\cos\theta&=\frac{3\sqrt{10}}{10}
.
\end{align}
Thus the sought area is
\begin{align}
S&=\tfrac13 \cdot 6^2 \cdot \tfrac13=4 
.
\end{align}
