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?

Let P, Q, R and S are the midpoint of AB, BC, CD and DA.
∴Co-ordinates of P = [x1+x2/2, y1+y2/2] = [-1-1/2, -1+4/2] = [-1, 3/2]
∴Co-ordinates of Q = [x1+x2/2, y1+y2/2] = [-1+5/2, 4+4/2] = [2, 4]
∴Co-ordinates of R = [x1+x2/2, y1+y2/2] = [5+5/2, 4-1/2] = [5, 3/2]
∴Co-ordinates of S = [x1+x2/2, y1+y2/2] = [5-1/2, -1-1/2] = [2, -1]
Now, length of PQ = √(-1-2)² + (3/2-4)² = √(-3)² + (-5/2)² = √9+25/4 = √61/4
Length of QR = √(2-5)² + (4-3/2)² = √(-3)² + (5/2)² = √9+25/4 = √61/4
Length of RS = √(5-2)² + (3/2+1)² = √(3)² + (5/2)² = √9+25/4 = √61/4
Length of SP = √(2+1)² + (-1-3/2)² = √(3)² + (-5/2)² = √9+25/4 = √61/4
Length of diagonal PR = √(-1-5)² + (3/2-3/2)² = √(-6)² = √36 = 6
Length of diagonal QS = √(2-2)² + (4+1)² = √(5)² = √25 = 5
Hence the all the sides of the quadrilateral PQRS are equal but the diagonals are not equal then PQRS is a rhombus.