posted by rohit

For what value of 'k' the points ( -k + 1, 2k), (k, 2 - 2k) and ( - 4 - k, 6 - 2k) are

  1. Steve

    they are collinear when the differences are all in the same ratio. That is, when the slope from P1 to P2 is the same as the slope from P2 to P3:

    ((2-2k)-(2k))/((k)-(-k+1)) = ((6-2k)-(2-2k))/((-4-k)-(k))
    (2-4k)/(2k-1) = (4)/(-4-2k)
    k = -1

    (2,-2),(-1,4),(-3,8): y=-2x+2

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