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In thea diagram,ABCD is a trapezium with AB,DC and vertices A(-4;1),B(x;-3),C(1;y) and D.Determine the value of x if the distance AB is 5 units.

  • geometry -

    A(-4,1), B(x,-3).
    (AB)^2 = (x-(-4))^2 + (-3-1)^2 = 5^2,
    (x+4)^2 + 16 = 25 - 16 = 9 ,
    Take sqrt of both sides:
    x+4 = +- 3,
    x = -4 +-3,
    x = -1, and x = -7.

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