Question

The equation of the bisector of the acute angles between the lines $3x-4y+7=0$ and $12x+5y-2=0$ is

• A. $99x-27y-81=0$
• B. $11x-3y+9=0$
• C. $21x+77y-101=0$
• D. $21x+77y+101=0$

Answer 1

Answer: (C)

$21x+77y-101=0$

The equations of the bisectors of the angles between two given lines are
$\frac{3x-4y+7}{\sqrt{3^2+4^2}}=\pm \frac{12x+5y-2}{\sqrt{{12}^2+5^2}}$
For bisector of acute angle $\frac{3x-4y+7}{5}=\frac{12x+5y-2}{13}$
$\Rightarrow 39x-52y+91=60x+25y-10$ or $21x+77y-101=0$