One bisector of the angle between the lines given by $a (x - 1)^{2} + 2 h (x - 1) y + b y^{2} = 0$ is 2x + y - 2 = 0.The other bisector is

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Solution

The correct option is D

We have $a (x - 1)^{2} + 2 h (x - 1) y + b y^{2} = 0$

or $a (x - 1)^{2} + 2 h (x - 1) (y - 0) + b (y - 0)^{2} = 0$

This equation represents a pair of straight lines intersecting at (1,0).Therefore shifting the origin at (1,0),we have x = X + 1, y = Y + 0 and the equation reduces to $a X^{2} + 2 h X Y + b Y^{2} = 0$ ..............(i)

One of the bisector of the angles between the lines given by (i) is 2x + y - 2 = 0 or 2(X + 1) + y - 2 = 0 i.e. 2X + Y = 0.Since the bisector are always at right angle , therefore the other bisector is X - 2Y = 0

i.e., x - 1 - 2y = 0 or x - 2y - 1 = 0.

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One bisector of the angle between the lines given by a(x−1)2+2h(x−1)y+by2=0 is 2x + y - 2 = 0.The other bisector is

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