STEP II 1992: Line Bisectors and Axes of Hyperbola I have been doing the problem given by
and I managed to successfully get to the last part, however,  I am stuck. I am not certain as to how to find the equation of the line bisectors. My first thought was to just change the axies to be the two lines find the bisector and then translate it backwards but that seems to be too much effort as it says "deduce". And for the last part (I assume axes means directix) I suspect the axes are going to be the same as the bisectors but I do not understand why. If anybody could help me with this I would greatly appreciate it.
 A: For the last part, the slopes of the two asymptotes $y=3x-4$ and $3y=x-4$ are 
$$ \tan a_1=3,\>\>\>\>\>\tan a_2= \frac13$$
which leads to $a_2= \frac\pi2-a_1$, i.e. the two asymptotes at the same angle with the $x$ and $y$ axes, respectively. 
Then, by symmetry, the angle bisectors are at 45-degree angle with the axes, and the slopes are $\pm 1$. Given that the asymptotes $y=3x-4$ and $3y=x-4$ meet at $(1,-1)$, the equations of the bisectors are $y+1=\pm(x-1)$, or,
$$y=x-2,\>\>\>\>\>y=-x$$
which are also the equations of the axes of the hyperbola.
A: The angle bisectors of a pair lines consist of all points that are equidistant from the lines. So, if you have the lines $ax+by+c=0$ and $px+qy+r=0$, use the standard point-line distance formula to construct the equation $${\lvert ax+by+c \rvert \over \sqrt{a^2+b^2}} = {\lvert px+qy+r \rvert \over \sqrt{p^2+q^2}}.$$ Removing the absolute value signs results in the two equations $${ax+by+c \over \sqrt{a^2+b^2}} = \pm{px+qy+r \over \sqrt{p^2+q^2}}$$ for the angle bisectors.  
Plugging in the two equations from the question, we have, after rearranging and eliminating the common factor of $\sqrt{10}$, $$(3x-y-4)+(x-3y-4) = 4x-4y-8 = 0$$ and $$(3x-y-4)-(x-3y-4) = 2x+2y=0.$$ The slope of the first line is $1$ and that of the second $-1$. You should be able to take it from here.
