A square in the coordinate plane has vertices whose $y$-coordinates are $0$, $1$, $4$, and $5$. What is the area of the square? $\textbf{(A)}\ 16\qquad$ $\textbf{(B)}\ 17\qquad$ $\textbf{(C)}\ 25\qquad$ $\textbf{(D)}\ 26\qquad$ $\textbf{(E)}\ 27$
Solution
The diagonals of a square bisect each other, so opposite vertices have the same sum of $y$-coordinates; hence the vertices with $y=0$ and $y=5$ are opposite. So we can take $A(0,0), B(b, 1), C(c, 5), D(d, 4)$.
Then $C=B+D-A$ gives $c=b+d$; $AB\perp AD$ gives $bd+4=0$; and $AB=AD$ gives $b^2+1=d^2+16$. So $b=4, d=-1, c=3$.
