Question

Find the equations of the bisectors of the angle between the straight lines $3x-4y+7=0$ and $12x-5y-8=0$

Answer 1

The equations of the bisectors of the angles between 3x - 4y + 7 = 0 and 12x - 5y - 8 = 0 are
$\frac{3x-4y+7}{\sqrt{3^2+{(-4)}^2}}=\pm \frac{12x-5y-8}{\sqrt{{12}^2+{(-5)}^2}}$
or $\frac{3x-4y+7}{5}=\pm \frac{12x-5y-8}{13}$
or $39x-52y+91=$ $\pm$ $(60x-25y-40)$
Taking the positive sign we get $21x+27y-131=0$ as one bisector
Taking the negative sign we get $99x-77y+51=0$ as the other bisector