Answer:
[tex]y = \frac{\sqrt{3}}{2}} \\\\[/tex]
Step-by-step explanation:
We are given a point P which lies on a unit circle
P = (-½, y)
Where x coordinate is -½ and y coordinate is y.
We are asked to find the value of y in simplest form.
Recall that a unit circle has a radius of 1 and is centered at the origin that is (0,0).
The equation of the unit circle is given by
[tex]x^{2} + y^{2} = 1[/tex]
Since the point P lies on a unit circle, we can substitute this point into the equation of unit circle and get the value of y.
Substitute x = -½ and y = y
[tex](-\frac{1}{2} )^{2} + y^{2} = 1\\\\(\frac{1}{4} ) + y^{2} = 1\\\\y^{2} = 1 - \frac{1}{4} \\\\y^{2} = \frac{3}{4} \\\\y = \pm \sqrt{\frac{3}{4}} \\\\y = \pm \frac{\sqrt{3}}{2}} \\\\[/tex]
Assuming that the given point P lies in the 2nd quadrant, the value of y will be positive.
[tex]y = \frac{\sqrt{3}}{2}} \\\\[/tex]
Therefore, the coordinates of the point P are
[tex]P =( -\frac{1}{2} , \frac{\sqrt{3}}{2}}) \\\\[/tex]