Question

The equations of the bisectors of the lines 3x-4y+7=0 and 12x-5y-8=0 are:

• A. 21x+27y-131=0
• B. 21x-27y+131=0
• C. 99x-77y+51=0
• D. 99x+77y-51=0

Answer 1

Answer: (A), (C)

99x-77y+51=0

The equations of the bisectors are :
$\frac{3x-4y+7}{\sqrt{3^2+{(-4)}^2}}=\pm \frac{12x-5y-8}{\sqrt{{12}^2+{(-5)}^2}}\Rightarrow \frac{3x-4y+7}{5}=\pm \frac{12x-5y-8}{13}\Rightarrow 39x-52y+91=\pm \left(60x-25y-40\right)\Rightarrow 21x+27y-131=0\mathrm{or}99x-77y+51=0$