# Find the equations of the bisectors of the angles between the straight lines $3x-4y+6=0$ and $12x+5y-3=0$.

$\begin{array}{1 1}(a)\;21x+77y-93=0,99x-77y+63=0\\(b)\;20x+77y-90=0,99x-77y+60=0\\(c)\;21x-77y+93=0,99x+77y+63=0\\(d)\;2x+70y-93=0,99x-70y+63=0\end{array}$

$\large\frac{(3x-4y+6)}{\sqrt{3^2+(-4)^2}}$$=\pm \large\frac{(12x+5y-3)}{\sqrt{(12)^2+(5)^2}} \large\frac{3x-4y+6}{5}=$$\pm \large\frac{(12x+5y-3}{13}$
$39x-52y+78=\pm(60x+25y-15)$
Taking +ve sign
$21x+77y-93=0$(one of the bisector)
Taking -ve sign
$99x-77y+63=0$
Hence (a) is the correct answer.