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 = 0`. Then other bisector is
(A) `2x-y=0`
(B) `2x+y=0`
(C) `2x+y-4=0`
(D) `x-2y+3=0`

Answer

1.5

Explanation

Shifting the origin to (1,2) , the equation of given pair of lines becomes
`ax^(2)+2hxy+by^(2)=0`
whose one angle bisector is `x+2y=0` (given) and other is `2x-y=0`.
`:.` Combined equation of angle bisector is
`(x+2y)(2x-y)=0`
On comparing with `(x^(2)-y^(2))/(a-b)=(xy)/(h)`, we get
`a-b=-3and h=2`
`:. (b-a)/(h)=(3)/(2)`