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Finding the inradius

  1. Nov 19, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    In ΔABC the cooridnates of the vertex A are(4,-1) and the lines x-y-1=0 and 2x-y=3 are internal bisectors of angles B and C then radius of incircle of ΔABC is

    2. Relevant equations

    3. The attempt at a solution
    I can solve the given two eqns to get intersection point which is (2,3). Finding the inradius requires me to calculate the distance from (2,3) to any side. For this I will need eqn of any one side. Now I can also find the 3rd bisector.
     
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  3. Nov 19, 2013 #2

    haruspex

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    There's probably a better way, but if the radius is r then you can write down the equation for a tangent from A to the circle. This will intercept the given lines at B and C.
     
  4. Nov 20, 2013 #3

    utkarshakash

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    Let us assume that the point of contact of tangent from A is (h,k). Equation of incircle:

    [itex](x-2)^2+(y-3)^2 = r^2 [/itex]

    Equation of tangent from A is [itex] y-k = \dfrac{2-h}{k-3} (x-h) [/itex]

    Since it passes through A, I get the following condition

    [itex]h^2+k^2-2k-6h+5=0[/itex]
     
  5. Nov 24, 2013 #4

    haruspex

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    Posting in the right thread this time:
    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?
     
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