Question

The equation of the bisector of the angle between the lines $3x-4y+7=0$ and $12x-5y-8=0$

• A. $99x-77y+51=0,21x+27y-131=0$
• B. $99x-77y+51=0,21x+27y+131=0$
• C. $99x-77y+131=0,21x+27y-51=0$
• D. None of these

Answer 1

The correct option is A $99x-77y+51=0,21x+27y-131=0$

Equation of the angle bisector of $3x-4y+7=0$ and $12x-5y-8=0$

is given by

$\frac{3x-4y+7}{\sqrt{3^2+4^2}}=\pm \frac{12x-5y-8}{\sqrt{{12}^2+5^2}}$

$⟹\frac{3x-4y+7}{5}=\pm \frac{12x-5y-8}{13}$

$\frac{3x-4y+7}{5}=\frac{12x-5y-8}{13}$

$39x-52y+91=60x-25y-40⟹21x+27y-131=0$ -----(1)

$\frac{3x-4y+7}{5}=-\frac{12x-5y-8}{13}$

$39x-52y+91=-60x+25y+40⟹99x-77y+51=0$ ---(2)

Therefore the equation of the bisector of the angle between $3x-4y+7=0$ and $12x-5y-8=0$ is $99x-77y+51=0$ and $21x+27y-131=0$