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`

One of the bisector of the angle between the lines `a(x-1)^2 + 2h(x-1)

1.	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» Correct Answer - 1.5 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)`

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