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Math Help - Angle bisector of a pair of straight lines

  1. #1
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    Angle bisector of a pair of straight lines

    hello everyone

    This is a question on coordinate geometry.
    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. We need to find the other bisector.

    Any help would be appreciated. Thanks in advance.
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  2. #2
    Super Member fardeen_gen's Avatar
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    Hello!


    We have a(x-1)^2 + 2h(x-1)y + by^2 = 0.

    Dividing throughout by (x - 1)^2, and solving the quadratic in \frac{y}{x - 1}, we get:
    \frac{y}{x - 1} = \frac{-h \pm \sqrt{h^2 - ab}}{b}, which gives the equations of the pair of straight lines.


    The joint equation of bisectors is given by:
    \frac{(x - 1)^2 - y^2}{a - b} = \frac{(x - 1)y}{h}
    (here x in the standard equation found in books is replaced by (x - 1) since the equation of our pair of straight lines is in (x - 1))


    On expanding, we get:

    hx^2 + xy(b - a) - hy^2 - 2hx + y(a - b) + h = 0 --- (1)

    One of the lines represented by the above equation is 2x + y - 2 = 0.

    We know that the bisectors of the angles between the lines are perpendicular to each other (Make sketches to convince yourself. Moreover in the joint equation, Coefficient of x^2 + Coefficient of y^2 = 0 which tells you this fact immediately).

    Thus, equation of the other bisector is x - 2y + k, where k is the constant to be determined.

    (We get the above equation by applying condition of perpendicularity: m_{1}m_{2} = -1, where m_{1} and m_{2} are the respective slopes. In our case m_{1} = -2 which gives m_{2} = \frac{1}{2}, thus giving the equation of the other bisector)

    Multiplying these two equations together,

    (2x + y  2)(x - 2y + k) = 0
    \Rightarrow 2x^2 - 3xy - 2y^2 + x(2k - 2) + y(4 + k) - 2k = 0

    Comparing with (1),
    1) h = 2
    2) 2k - 2 = -2h \Rightarrow k = -1

    So, the other bisector is:
    \boxed{x - 2y - 1 = 0}
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  3. #3
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    thanks fardeen, i couldnt have asked for more.
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  4. #4
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    Quote Originally Posted by watsmath View Post
    hello everyone

    This is a question on coordinate geometry.
    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. We need to find the other bisector.

    Any help would be appreciated. Thanks in advance.
    Alternatively,

    we can find the point of intersection of the angle bisector 2x+y-2 with the lines
    and find the equation of the perpendicular to the bisector that goes through
    that same point of intersection.

    Hence, y=2-2x is substituted

    a(x-1)^2+2h(x-1)(2-2x)+b(2-2x)^2=0

    2-2x=2(1-x)=-2(x-1)

    (2-2x)^2=2(1-x)2(1-x)=-2(x-1)(-2)(x-1)=4(x-1)^2

    then

    a(x-1)^2-4h(x-1)+4(x-1)^2=0

    (x-1)\left[a(x-1)-4h+4(x-1)\right]=0

    x=1

    2x+y-2=0\ \Rightarrow\ 2+y-2=0,\ y=0

    The slope of 2x+y-2=0 is -2, as y=-2x+2

    Any perpendicular line has slope \frac{1}{2}

    The perpendicular bisector contains the point (1,0)

    hence it's equation is

    y-0=\frac{1}{2}(x-1)

    2y=x-1

    x-2y-1=0
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