Question

The straight line $2x+3y+1=0$ bisects the angle between two straight lines one of which is $3x+2y+4=0$. Determine the equation of the other line.

Answer 1

The line $l_2$ will pass through the intersection of $l_1=0$ and $B=0$. Its equation is
$3x+2y+4+\lambda (2x+3y+!)=\mathrm{0....}\left(1\right)$
Now choose a point $(\alpha ,\beta )$ on the bisector so that $2\alpha +3\beta +1=0$. Apply the condition $p_1=p_2$ for the two lines
$\frac{3\alpha +2\beta +4+\lambda (2\alpha +3\beta +1)}{{(3+2\lambda )}^2+{(2+3\lambda )}^2}=\frac{3\alpha +2\beta +4}{\sqrt{9+4}}$
Put $2\alpha +3\beta +1=0$. Cancel $3\alpha +2\beta +4$
$\therefore 13=13{\lambda }^2+24\lambda +13=0$
$\therefore \lambda =0,-\frac{24}{13}$
But $\lambda =0$ corresponds to given line hence $\lambda =-\frac{24}{13}$ will give the other line whose equation from (1) is $9x+46y=28$