Find the bisector of the angles between the pair of straight lines $4x + 3y = 3$ and $5x - 12y = 8$, placing first the bisector of that angle in which the origin lies.
Equations of the given lines are: $\;\;$ $4x + 3y = 3$, $\;\;\;$ $5x - 12y = 8$
Writing the equations so that their constant terms are negative, they are
$4x + 3y -3 = 0$, $\;\;\;$ $5x - 12y - 8 = 0$
$\therefore \;$ The equation to the bisector of the angle in which the origin lies is
$\dfrac{4x + 3y - 3}{\sqrt{4^2 + 3^2}} = \dfrac{5x - 12y - 8}{\sqrt{5^2 + 12^2}}$
i.e. $\;$ $\dfrac{4x + 3y - 3}{5} = \dfrac{5x - 12y - 8}{13}$
i.e. $\;$ $13 \left(4x + 3y - 3\right) = 5 \left(5x - 12y - 8\right)$
i.e. $\;$ $52 x + 39 y - 39 = 25 x - 60 y - 40$
i.e. $\;$ $27 x + 99y + 1 = 0$
The equation to the other bisector is
$\dfrac{4x + 3y - 3}{\sqrt{4^2 + 3^2}} = \dfrac{- \left(5x - 12y - 8\right)}{\sqrt{5^2 + 12^2}}$
i.e. $\;$ $\dfrac{4x + 3y - 3}{5} = \dfrac{- \left(5x - 12y - 8\right)}{13}$
i.e. $\;$ $13 \left(4x + 3y - 3\right) = -5 \left(5x - 12y - 8\right)$
i.e. $\;$ $52 x + 39 y - 39 = -25 x + 60 y + 40$
i.e. $\;$ $77 x - 21y - 79 = 0$