equations of triangle's sides, given equations of two bisectors and one point of triangle? The equations two angles bisectors are $x-3y-6=0$ and $x+y-2=0$. We also know that one point of the triangle is $A(2,-4)$. Clearly, this point doesn't satisfy those two equations, and by finding the intersection of the bisectors and A we can write the third bisector's equation. This is all I can come up with. Any hints?
 A: Hint 
Take the image of Vertex $A$ in the two bisector to get two points.  Write the equation of line of these points and solve it with the equation of bisectors to get the other vertices $B$ and $C$ respectively.
A: Let $D$ be the incentre. $D=(3,-1)$.
The slope of $AD$ is $3$.
Let $\angle BAC=2a$, $\angle ABC=2b$ and $\angle ACB=2c$. Then $a+b+c=\frac{\pi}{2}$.
Note that $\angle BDC=\pi-b-c=\frac{\pi}{2}+a$.
If $\theta$ is the acute angle between the two given angle bisectors, then
$$\tan\theta=\frac{\frac{1}{3}-(-1)}{1+(-1)(\frac{1}{3})}=2$$
As $\frac{\pi}{2}+a$ is obtuse, $\frac{\pi}{2}+a=\pi-\theta$ and hence $a=\frac{\pi}{2}-\theta$. $\tan a=\frac{1}{2}$.
Let $m$ be the slope of a line making an angle $a$ with $AD$.
\begin{align*}
\tan a&=\left|\frac{m-3}{1+3m}\right|\\
\frac{1}{2}&=\left|\frac{m-3}{1+3m}\right|\\
1+3m&=\pm2(m-3)\\
m&=1\quad\textrm{or}\quad-7
\end{align*}
The equations of two of the sides are $y=-7x+10$ and $y=x-6$.
We can find $B$ and $C$ by solving these two lines with the angle bisectors.
