2014 AMC 10A Problem 18

Opposite vertices of a square have the same average $y$-coordinate. Thus the opposite pairs must have $y$-coordinates $(0,5)$ and $(1,4)$, the two pairs with equal sum. In particular, a vertex with $y$-coordinate $0$ is adjacent to vertices with $y$-coordinates $1$ and $4$. Let one vertex be $A=(0,0)$, and let its adjacent vertex with $y$-coordinate $1$ be $B=(x,1)$ with $x>0$. Rotating the side vector $(x,1)$ by $90^\circ$ gives the next side vector $(-1,x)$, so another vertex has y-coordinate $x$. The remaining y-coordinates are $4$ and $5$, hence $x=4$. The side length squared is $AB^2=x^2+1^2=4^2+1=17$, which is the area of the square. Thus, B is the correct answer.