Answer: All four sides equal \(3\sqrt{10}\), so ABCD is a rhombus.
Distance formula: \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
\[AB=\sqrt{3^2+9^2}=\sqrt{90}=3\sqrt{10}\]
\[BC=\sqrt{9^2+3^2}=\sqrt{90}=3\sqrt{10}\]
\[CD=\sqrt{(-3)^2+(-9)^2}=\sqrt{90}=3\sqrt{10}\]
\[DA=\sqrt{(-9)^2+(-3)^2}=\sqrt{90}=3\sqrt{10}\]
A quadrilateral with all sides equal is a rhombus. (Its diagonals \(AC=\sqrt{288}\) and \(BD=\sqrt{72}\) are unequal, so it is not a square.)