Question

The equation of the bisector of the lines 4x+3y-6=0 and 5x+12y+9=0 containing the origin is given by .

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

Answer 1

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

We firstly write the equations of lines in such form where signs of constant terms is same (let's say positive).
$\Rightarrow$ -4x-3y+6=0 and 5x+12y+9=0
Now, the equation of the bisector containing the origin will be the one corresponding to the positive sign.
$\Rightarrow \frac{-4x-3y+6}{\sqrt{(-4{)}^2}+\sqrt{(-3)}}=+\frac{5x+12y+9}{\sqrt{5^2}+\sqrt{{12}^2}}\Rightarrow \frac{-4x-3y+6}{5}=\frac{5x+12y+9}{13}\Rightarrow -52x-39y+78=25x+60y+45\Rightarrow 7x+9y-3=0$