geometry
posted by Phillip .
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.

A(4,1), B(x,3).
(AB)^2 = (x(4))^2 + (31)^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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