✔ 最佳答案
As is a point on the y-axis,
Let the coordinates of U be (0,y)
SU = TU (given)
√[ (y﹣8)2 + (0﹣4)2 ] = √{ [y﹣(-4)]2 + [0﹣(-2)2] }
√[ (y﹣8)2 + 16 ] = √[(y + 4)]2 + (0 + 2)2]
(y﹣8)2 + 16 = (y + 4)2 + 4
y2﹣16y + 64 + 16 = y2 + 8y + 16 + 4
60 = 24y
y = 5 / 2
∴the coordinates of U is (0, 5/2)
2008-07-30 15:50:54 補充:
If a point is on the y-axis,then its x-coordinates should be 0.