Question

For the straight lines $4x+3y–6=0$ and $5x+12y+9=0$, find the equation of the bisector of the angle which contains $(1,2)$

Answer 1

For the point $(1,2),4x+3y–6=4\times 1+3\times 2–6>0$

$5x+12y+9=12\times 2+9>0$

If $\theta$ is the acute angle between the line $4x+3y–6=0$ and the bisector $9x–7y–41=0,$ then

Hence equation of the bisector of the angle containing the point $(1,2)$ is

$\frac{4x+3y-6}{5}=\frac{5x+12y+9}{13}$

$\Rightarrow 52x+39y-78=25x+60y+45$

or $27x-21y-123=0$

or $9x–7y–41=0$