Answer: All sides equal √5 and both diagonals equal √10, so ABCD is a square.
Distance formula: \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
\(AB=\sqrt{4+1}=\sqrt5\), \(BC=\sqrt{1+4}=\sqrt5\), \(CD=\sqrt{4+1}=\sqrt5\), \(DA=\sqrt{1+4}=\sqrt5\).
Diagonals: \(AC=\sqrt{9+1}=\sqrt{10}\), \(BD=\sqrt{1+9}=\sqrt{10}\).
All sides equal makes ABCD a rhombus; a rhombus with equal diagonals is a square.