Question

The line $lx+my+n=0$ bisects the angle between a pair of straight lines of which one is $px+qr+r=0$. Show that the equation to the other line is
$(px+qy+r)(l^2+m^2)-2p(lp+mq)(lx+my+n)=0$

Answer 1

$lx+mu+n=0$ is a bisector and let $(\alpha ,\beta )$ be any point on it so that
$l\alpha +m\beta +n\gamma =\mathrm{0....}\left(1\right)$
the other line will pass through the intersection of given lines and given bisector and hence by $P+\lambda Q=0$ its equation is
$(px+qy+r)+\lambda (l\alpha +m\beta +n\gamma )=\mathrm{0....}\left(2\right)$
Also $px+qy+r=\mathrm{0......}\left(3\right)$
If $(\alpha ,\beta )$ be a point on the bisector then its perpendicular distance from the lines (2) and (3) is same
$\therefore \frac{(p\alpha +q\beta +r)+\lambda (l\alpha +m\beta +n\gamma )}{\sqrt{{(p+l\lambda )}^2+{(q+m\lambda )}^2}}=\pm \frac{p\alpha +q\beta +r}{\sqrt{p^2+q^2}}$
Putting $l\alpha +m\beta +n\gamma =0$ by (1) in the above and cancelling $p\alpha +q\beta +r$ and then squaring both sides we get
${(p+l\lambda )}^2+{(q+m\lambda )}^2=p^2+q^2$
or $2\lambda (pl+qm)+{\lambda }^2(l^2+m^2)=0$
$\therefore \lambda =-2.\frac{pl+qm}{l^2+m^2}$
Putting $\lambda$ in (2) the required line is
$(l^2+m^2)(px+qy+r)-2(lp+mq)(lx+my+n)=0$