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)

$99x + 27y+ 30= 0$

Answer 2

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