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coordinate geometry

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Find the value of x such that PQ=QR where the coordinates of P,Q and R are [6,-1], [1,3]and[x,8] respectively

  • coordinate geometry -

    PQ2 = 52 + 42
    QR2 = (x-1)2 + 5^2

    Looks like x-1 = 4 or x=5

  • coordinate geometry - PS -

    or, x-1 = -4 making x=-3

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