Answer
$P(- \frac{\sqrt 5}{3},\frac{2}{3})$
Work Step by Step
We have the following equation for the unit circle:
$x^{2}+y^{2}=1$, so we plug in $P(x,\frac{2}{3})$,
$(\frac{2}{3})^{2}+x^{2}=1$
$ x^{2}=1 - \frac{4}{9}= \frac{5}{9}$
$x = \frac{\sqrt 5}{3}$ or $x = - \frac{\sqrt 5}{3}$ but the x-coordinate is negative so $x = - \frac{\sqrt 5}{3}$
