Question

Equation of the bisector of the acute angle between lines $3x+4y+5=0$ and $12x-5y-7=0$ is

• A. $21x+77y+100=0$
• B. $99x-27y+30=0$
• C. $99x+27y+30=0$
• D. $21x-77y+100=0$

Answer 1

Answer: (C)

Given equations are
$3x+4y+5=0$ and $12x-5y-7=0$
$\therefore a_1{a}_2+b_1{b}_2=3\times 12+4\times (-5)$
$=16>0$
$\therefore$ For acute angle bisector
$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=-\frac{\left(a_{2}x+b_{2}y+c_2\right)}{\sqrt{a_2^2+b_2^2}}$
$\therefore \frac{3x+4y+5}{\sqrt{9+16}}=-\frac{(12x-5y-7)}{\sqrt{12^2+(-5{)}^2}}$
$\Rightarrow \frac{3x+4y+5}{5}=-\frac{(12x-5y-7)}{13}$
$\Rightarrow 39x+52y+65=-60x+25y+35$
$\Rightarrow 99x+27y+30=0$