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 the origin

Answer 1

For point O(0, 0) $4x+3y-6=-6<0$ and $5x+12y+9=9>0$
Hence for point O(0, 0) $4x+3y-6$ and $5x+12y+9$ are of opposite signs
Hence equation of the bisector of the angle between the given lines containing the origin will be
$\frac{4x+3y-6}{\sqrt{{\left(4\right)}^2+{\left(3\right)}^2}}=-\frac{5x+12y+9}{\sqrt{5^2+{12}^2}}$
or $\frac{4x+3y-6}{5}=-\frac{5x+12y+9}{13}$ or $52x+39y-78=-25x-60y-45$
or $77x+99y-33=0or7x+9y-3=0$