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Finding angle bisectors

  1. Nov 19, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    One bisector of the angle between the lines given by [itex]a(x-1)^2 + 2h(x-1)y + by^2=0[/itex] 2x+y-2=0. The other bisector is

    3. The attempt at a solution
    Expanding
    [itex]ax^2+2hxy+by^2-2ax-2hy+a=0[/itex]

    The angle between the pair of lines is given by

    [itex]|\dfrac{2\sqrt{h^2-ab}}{a+b}|[/itex]

    But those constants are still unknown to me. I can't really figure out how the given bisector can help me?
     
  2. jcsd
  3. Nov 19, 2013 #2
    I suggest shifting the origin to ##(1,0)## and convert the given equation to a homogeneous second degree equation.

    Can you take it from here?
     
  4. Nov 19, 2013 #3

    utkarshakash

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    Shifting the origin only makes the equation simpler. The basic question still remains the same and this is where I'm getting stuck.
     
  5. Nov 19, 2013 #4
    I am sorry to mislead you. You don't really need to shift the origin.

    What do you think is the angle between the angle bisectors?
     
  6. Nov 20, 2013 #5

    utkarshakash

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    I have already posted the formula needed to calculate it in my original post. However I don't know the constants.
     
  7. Nov 20, 2013 #6
    The formula you quote is for the angle between the given pair of line, not the angle bisectors.

    And you don't need a formula for angle between the angle bisectors. Think simple. Draw two intersecting lines, draw their angle bisectors and find the angle between them. Its quite straightforward then.
     
  8. Nov 20, 2013 #7

    utkarshakash

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    Both will be perpendicular to each other. But what about the intersection point? I only know that slope of other bisector will be 1/2. I also need to find a point to find the exact equation.
     
  9. Nov 20, 2013 #8

    haruspex

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    Where does the pair of lines given by the quadratic intersect? You have the equations for those two lines, right?
     
  10. Nov 23, 2013 #9

    haruspex

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    Try this:
    Drop a perpendicular from the incentre O to each of the three sides, meeting AB at C', BC at A' and CA at B'. ∠AOB' = ∠AOC' = α, say; ∠BOA' = ∠BOC' = β, etc. So what does α+β+γ equal? What is ∠BOC in terms of these? You can determine the value of ∠BOC. What does that give for the value of α? Can you use AO and α to find r?
     
  11. Nov 24, 2013 #10

    utkarshakash

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    I think you mistook this post for my another post. However, your method was really a good one.
     
  12. Nov 24, 2013 #11

    haruspex

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    Ah - yes. Just copied it to the right thread for the benefit of anyone else reading them.
     
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