If the point P(6/7,y) is on the unit circle in quadrant IV, then y=

Answer 1

We are in the 4th quadrant so y will be negative. r is always taken to be positive and in this case it is 1 because it is a unit circle.

(6/7)^2 + y^2 = 1^2

(36/49) + y^2 = 1

y^2 = 1 - 36/49

y^2 = 13/49

y = +/- (sqrt13)/7 so in this case y = - (sqrt13)/7