Question

The equation of the obtuse angle bisector of lines $12x-5y+7=0$ and $3y-4x-1=0$ is

• A. $4x+7y+11=0$
• B. $4x+7y-11=0$
• C. $7x+4y+11=0$
• D. $7x+4y-11=0$

Answer 1

The correct option is A $4x+7y+11=0$

Given lines $12x-5y+7=0$ and $3y-4x-1=0$
Now, making the constant terms positive, we get
$12x-5y+7=0$ and $4x-3y+1=0$
Checking
$a_1{a}_2+b_1{b}_2=48+15=63>0$
So, the obtuse angle bisector is
$\frac{12x-5y+7}{\sqrt{(12{)}^2+(-5{)}^2}}=+\frac{4x-3y+1}{\sqrt{(4{)}^2+(-3{)}^2}}\Rightarrow \frac{12x-5y+7}{13}=\frac{4x-3y+1}{5}\Rightarrow 60x-25y+35=52x-39y+13\Rightarrow 8x+14y+22=0\therefore 4x+7y+11=0$