Find the equation of the line that bisects the acute angles determined by the given lines.

1) $\displaystyle x-6y-12=0 $ and $\displaystyle 4x-6y-9=0 $

Attempt:

$\displaystyle d1=\frac{x-6y-12}{-\sqrt{37}} $ , $\displaystyle \frac{4x-6y-9}{-2\sqrt{13}} $

So d1+d2 =0

$\displaystyle d1=\frac{x-6y-12}{-\sqrt{37}} $ +$\displaystyle \frac{4x-6y-9}{-2\sqrt{13}} $

=$\displaystyle \frac{(-4\sqrt{37} -2\sqrt{13} )x +(6\sqrt{37} +12\sqrt{13} )y+(-9\sqrt{37} -24\sqrt{13} )}{2\sqrt{481}} $

How to make the equation free of square roots?

And is my solution correct ( if not please follow the procedure that i have done which is d1+d2=0 )

Thank you.