If the point P(6/8,y) is on the unit circle in quadrant IV, then y=

Answer 1

y^2 = 1 - (6/8)^2 = 1 - (3/4)^2 = 1- 9/16 = 7/16

y = sqr(7/16) = -(1/4)sqr7. ignore the positive square root as y is only negative in quadrant 4

y = sqr7/4