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Thread: 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 $\displaystyle a(x-1)^2 + 2h(x-1)y + by^2 = 0$.

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


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


    On expanding, we get:

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

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

    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 $\displaystyle x^2$ + Coefficient of $\displaystyle y^2$ = 0 which tells you this fact immediately).

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

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

    Multiplying these two equations together,

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

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

    So, the other bisector is:
    $\displaystyle \boxed{x - 2y - 1 = 0}$
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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

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

    $\displaystyle 2-2x=2(1-x)=-2(x-1)$

    $\displaystyle (2-2x)^2=2(1-x)2(1-x)=-2(x-1)(-2)(x-1)=4(x-1)^2$

    then

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

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

    x=1

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

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

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

    The perpendicular bisector contains the point (1,0)

    hence it's equation is

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

    $\displaystyle 2y=x-1$

    $\displaystyle x-2y-1=0$
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