$ABCD$ is a rectangle formed by the points $A(-1, -1)$,$B(-1, 4)$,$C(5, 4)$,and $D(5, -1)$. $P, Q, R$,and $S$ are the mid-points of $AB, BC, CD$,and $DA$ respectively. Is the quadrilateral $PQRS$ a square,a rectangle,or a rhombus? Justify your answer.

(C) $P$ is the mid-point of side $AB$. Therefore,the coordinates of $P$ are $\left(\frac{-1-1}{2}, \frac{-1+4}{2}\right) = \left(-1, \frac{3}{2}\right)$.
Similarly,the coordinates of $Q, R$,and $S$ are $(2, 4)$,$\left(5, \frac{3}{2}\right)$,and $(2, -1)$ respectively.
Length of $PQ = \sqrt{(-1-2)^2 + \left(\frac{3}{2}-4\right)^2} = \sqrt{(-3)^2 + \left(-\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}}$.
Length of $QR = \sqrt{(2-5)^2 + \left(4-\frac{3}{2}\right)^2} = \sqrt{(-3)^2 + \left(\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}}$.
Length of $RS = \sqrt{(5-2)^2 + \left(\frac{3}{2}-(-1)\right)^2} = \sqrt{3^2 + \left(\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}}$.
Length of $SP = \sqrt{(2-(-1))^2 + \left(-1-\frac{3}{2}\right)^2} = \sqrt{3^2 + \left(-\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{61}{4}}$.
Length of diagonal $PR = \sqrt{(-1-5)^2 + \left(\frac{3}{2}-\frac{3}{2}\right)^2} = \sqrt{(-6)^2 + 0^2} = 6$.
Length of diagonal $QS = \sqrt{(2-2)^2 + (4-(-1))^2} = \sqrt{0^2 + 5^2} = 5$.
Since all sides are equal $(PQ = QR = RS = SP = \sqrt{\frac{61}{4}})$ and the diagonals are not equal $(PR \neq QS)$,the quadrilateral $PQRS$ is a rhombus.