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