find acute bisector between 2 lines

3x+4y=11,12x-5y=2

3x + 4y = 11, 12x - 5y = 2
Here we have to make the constant term positive.
So -3x -4y +11 = 0 and -12x + 5y +2 = 0
Now we have to check a₁a₂ + b₁b₂
So a₁a₂ + b₁b₂  = 36 -20 = 16 > 0
So the equation of bisector between two lines is
$$\begin{gathered} \frac{-3x -4y +11}{\sqrt{9+16}} =-\frac{-12x +5y +2}{\sqrt{144+25}} \\ \Rightarrow 13(-3x-4y+11) =-5(-12x + 5y +2) \\ \Rightarrow 11x +3y -17 = 0 \\ Hence this is the required equation of bisector. \end{gathered}$$