geometry
posted by fakku .
The distance between (3,1) and (3,y) is sq rt45.Find y?

√((3+3)^2 + (y+1)^2 ) = √45
square both sides
36 + (y+1)^2 = 45
(y+1)^2 = 9
y+1 = ± 3
y = 1 ± 3
= 2 or 4
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