equations of triangle's sides, given equations of two bisectors and one point of triangle?

Question

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?

Answer 1

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.

Answer 2

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.