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Math Help - intersection of tangents to a curve, equidistance

  1. #1
    Senior Member furor celtica's Avatar
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    intersection of tangents to a curve, equidistance

    It is given that y = 0.5(x^2) – 3x + 12
    The points P and Q on the graph have x-coordinates 0 and 8 respectively. The tangents at P and Q meet at R. Show that the point (11, 9) is equidistant from P, Q and R.

    Alright so this is my work so far:
    dy/dx = 2x – 3
    P (0, 12), Q (8, 20)
    Equation of tangent at P: (y – 12)/(x – 0) = -3 => y = 12 - 3x
    Equation of tangent at Q: (y – 20)/(x – 8) = 13 => y = 13x - 84
    Point of intersection R is defined by: 12 – 3x = 13x – 84 => 16x = 96 => x = 6
    R(6,-6)

    Distance between (11, 9) and P = sqrt((11^2) + (-3^2)) = sqrt(130)
    Distance between (11, 9) and Q = sqrt((3^2) + (-11^2)) = sqrt(130)
    Distance between (11, 9) and R = sqrt(((11-6)^2) + ((9+6))^2) = sqrt(250)

    So there’s my problem. Where have I messed up?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: intersection of tangents to a curve, equidistance

    The problem is your derivative, it has to be:
    \frac{d}{dx}\left(\frac{1}{2}x^2-3x+12\right)=\frac{1}{2}\cdot 2x-3=x-3
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