Trigonometry 7th Edition

$(\frac{\sqrt 2}{2}$,$\frac{\sqrt 2}{2}$), $(-\frac{\sqrt 2}{2}$,$-\frac{\sqrt 2}{2}$)
Since $y=x$ and $x^2+y^2=1$, the first thing that we have to do is substitute, as follows: $y=x$ $x^2+x^2=1$. The next step would be to solve for x as follows: $2x^2=1$ $x^2=\frac{1}{2}$ $x=± \sqrt \frac{1}{2}$ $x= ±\frac{\sqrt 2}{2}$. The following step would be to find y as follows. Since $y=x$ and $x=±\frac{\sqrt 2}{2}$, then $y=±\frac{\sqrt 2}{2}$. Therefore, the final answer is $(\frac{\sqrt 2}{2}$,$\frac{\sqrt 2}{2}$), $(-\frac{\sqrt 2}{2}$,$-\frac{\sqrt 2}{2}$).