What is the shortest method to solve this sum?
One of the bisector of the angle between the lines $a(x-1)^2+2h(x-1)(y-2)+b(y-2)^2=0$ is $x+2y=5$.The other bisector is what?
My approach is becoming long.I took the other bisector as mx+ny+p=0 and then compared $(mx+ny+p)(x+2y-5)=0$ with the joint equation of bisectors.
The other bisector is perpendicular to the first, so you can save yourself some calculation by fixing $m$ and $n$ so this is the case.