Answer:So first, I found the length of the sides and the diagonal of the square, which are 18−−√ and 6 respectively. By graphing, I know the solution is (0,−1). Then, I assume that since the length between (3,2) and (−3,2) is the diagonal, then the distance between (0,5) and the remaining vertex must be the diagonal too. And since the length of the side is 6, then the distance between the vertex and either (3,2) or (−3,2) must be 6. So:
(x−3)2+(y−2)2−−−−−−−−−−−−−−−√=18−−√
(x−0)2+(y−5)2−−−−−−−−−−−−−−−√=6
Which gives (after a bit of cleaning up):
x2+y2−10y=11
x2−6x+y2−4y=5
Then, replacing the second expression into the first one:
x2−6x+y2−4y=5⇒x2=5+6x−y2+4y
5+6x−y2+4y+y2−10y=11
5+6x+4y−10y=11
6x−6y+6
x−y=1
x=1+y
Up to this point, I know I'm not entirely wrong because the expression is true for the actual coordinates of the vertex, because 0=1+(−1) is true. But I wouldn't know how to proceed if I hadn't known the answer beforehand. I need to find both x and y, is there a linear equation I'm missing to find the exact coordinates of the last vertex? Is my process okay or is there a simpler way to do it?
Step-by-step explanation:
