Question

The equation(s) of the angle bisectors of the lines $3x-4y+7=0$ and $12x-5y-8=0$ is/are

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

Answer 1

Answer: (C), (D)

The correct options are
C $99x-77x+51=0$
D $21x+27y-131=0$

Given lines are $3x-4y+7=0$ and $12x-5y-8=0$
We know that the equation of angle bisectors is
$\frac{3x-4y+7}{\sqrt{3^2+(-4{)}^2}}=\pm \frac{12x-5y-8}{\sqrt{(12{)}^2+(-5{)}^2}}\Rightarrow \frac{3x-4y+7}{5}=\pm \frac{12x-5y-8}{13}$
$\Rightarrow 39x-52y+91=\pm (60x-25y-40)$
Now, the two bisectors are
$39x-52y+91=60x-25y-40\Rightarrow 21x+27y-131=0$
And
$39x-52y+91=-(60x-25y-40)\Rightarrow 99x-77x+51=0$