Answer
$(\frac{\sqrt 2}{2}$,$\frac{\sqrt 2}{2}$), $(-\frac{\sqrt 2}{2}$,$-\frac{\sqrt 2}{2}$)
Work Step by Step
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
$(\frac{\sqrt 2}{2}$,$\frac{\sqrt 2}{2}$), $(-\frac{\sqrt 2}{2}$,$-\frac{\sqrt 2}{2}$).
