Question

The straight lines $3x-4y+7=0$ and $12x+5y-2=0$ has equation of the acute angle bisector as:

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

Answer 1

Answer: (B)

Equation of the straight lines are
$3x-4y+7=\mathrm{0..........}(i)$
and $-12x-5y+2=\mathrm{0.............}(ii)$
Here $(3)(-12)+(-4)(-5)+(7)(2)$
$=-36+20+14=-2<0$
$\therefore$ it makes acute angle.
$\therefore$ equation of the bisector of an acute angle between these lines is
$\frac{3x-4y+7}{\sqrt{3^2+4^2}}=+\frac{\left(-12x-5y+2\right)}{\sqrt{{\left(12\right)}^2+{\left(5\right)}^2}}$ .
$\Rightarrow 99x-27y+81=0$
$\Rightarrow 11x-3y+9=0$