Question

The equation of the bisector of the acute angle between $4x+3y-6=0$ and $5x+12y+9=0$ is

• A. $9x-7y-41=0$
• B. $7x+9y-3=0$
• C. $9x+7y-3=0$
• D. $9x-7y+41=0$

Answer 1

The correct option is B $7x+9y-3=0$

Given lines are $4x+3y-6=0$ and $5x+12y+9=0$
Making the constant terms positive, we get
$-4x-3y+6=0$ and $5x+12y+9=0$
Now,
$a_1{a}_2+b_1{b}_2=-20-36=-56<0$
So, the equation of acute angle bisector is
$\frac{-4x-3y+6}{\sqrt{(-4{)}^2+(-3{)}^2}}=\frac{5x+12y+9}{\sqrt{(5{)}^2+(12{)}^2}}\Rightarrow \frac{-4x-3y+6}{5}=\frac{5x+12y+9}{13}$
$\Rightarrow 13(-4x-3y+6)=5(5x+12y+9)\Rightarrow -52x-39y+78=25x+60y+45\Rightarrow 77x+99y-33=0\therefore 7x+9y-3=0$