One bisector of the angle between the lines g | vedclass

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(D) The equation $a(x - 1)^2 + 2h(x - 1)y + by^2 = 0$ represents a pair of straight lines intersecting at the point $(1, 0)$.By shifting the origin to

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$x - 2y - 1 = 0$

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(D) The equation $a(x - 1)^2 + 2h(x - 1)y + by^2 = 0$ represents a pair of straight lines intersecting at the point $(1, 0)$.By shifting the origin to $(1, 0)$,we substitute $x = X + 1$ and $y = Y$,which transforms the equation into $aX^2 + 2hXY + bY^2 = 0$ $(i)$.The given bisector $2x + y - 2 = 0$ becomes $2(X + 1) + Y - 2 = 0$,which simplifies to $2X + Y = 0$.Since the angle bisectors of a pair of straight lines are always perpendicular to each other,the slope of the second bisector must be the negative reciprocal of the slope of the first bisector.The slope of $2X + Y = 0$ is $m_1 = -2$. Thus,the slope of the second bisector $m_2$ is $1/2$.The equation of the second bisector passing through $(1, 0)$ is $Y - 0 = \frac{1}{2}(X - 0)$,which simplifies to $X - 2Y = 0$.Substituting back $X = x - 1$ and $Y = y$,we get $(x - 1) - 2y = 0$,or $x - 2y - 1 = 0$.

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

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