Test whether the point is on the unit circle by plugging it into the equation, x 2 + y 2 = 1 x^2+y^2=1 x 2 + y 2 = 1 . ( − 2 2 , − 2 2 ) \left(\frac{-\sqrt2}{2},\frac{-\sqrt2}{2}\right) ( 2 − 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> , 2 − 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> )

In this problem we're told to test whether the point is on the unit circle by plugging it into the equation, and our unit circle equation is right here x squared plus y squared equals one. So let's go ahead and do that. Now looking at this equation x squared plus y squared equals one, and the point that I'm given, I have my x value and my y value, we can just go ahead and plug those right in. Now we're going to check if it makes this equation true because if it does, we know that this point is going to be located somewhere on our unit circle. So we're testing, does x squared plus y squared equal one by plugging in these values. So we have negative root two over two and we want to square that and we're adding that together with again root negative root two over two because they happen to have the same values here. Squaring that and we want to know is this equal to one. So let's go ahead and do some algebra here. Now squaring this fraction I need to make sure to distribute that square to both the numerator and the denominator. So squaring the numerator that negative will go away because negative one squared is a positive. So this will give me two because I'm really just canceling out that square root. And then in my denominator, I have a value of four because I'm squaring that value of two. Now because these x and y values are the exact same, I know that my y squared will be the same thing. So I have two fourths plus two fourths and I'm still testing is this equal to one. Now looking at these adding them together they already have the same denominator so this gives me four over four just adding those numerators together and we're checking is this equal to one. Now my knowledge of fraction kicks in here. I know that four over four is indeed one, so this is really just saying that one is equal to one, which is definitely a true statement. So we know that, yes, this is on the unit circle located somewhere on that. So thanks for watching, and let me know if you have questions.