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 midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
P id the mid-point of side AB,
Coordinate of P = \(\frac{(-1-1)}{2}, \frac{(-1+4)}{2}\) = \((-1), \frac{3}{2}\)
Similarly, Q, R and S are (As Q is mid-point of BC, R is midpoint of CD and S is midpoint of AD)
Coordinate of Q = (2, 4)
Coordinate of R = (5, 3/2)
Coordinate of S = (2, -1)
Length of PQ = \(\sqrt{(-1-2)^2 + (\frac{3}{2}-4)^2}\)
= \(\sqrt{\frac{61}{4}}\)
= \(\sqrt{\frac{61}{2}}\)
Length of SP = \(\sqrt{(2+1)^2 + (-1 - \frac{3}{2})^2}\)
= \(\sqrt{\frac{61}{4}}\)
= \(\sqrt{\frac{61}{2}}\)
Length of QR = \(\sqrt{(2-5)^2 + (4 - \frac{3}{2})^2}\)
= \(\sqrt{\frac{61}{4}}\)
= \(\sqrt{\frac{61}{2}}\)
Length of RS = \(\sqrt{(5-2)^2 + (4 - \frac{3}{2})^2}\)
= \(\sqrt{\frac{61}{4}}\)
= \(\sqrt{\frac{61}{2}}\)
Length of PR (diagonal) =
\(\sqrt{(-1-5)^2 + (\frac{3}{2} - \frac{3}{2})^2}\)
= 6
Length of QS (diagonal) = \(\sqrt{(2-2)^2 + (4 + 1)^2}\) = 5
The above values show that, PQ = SP = QR = RS = \(\sqrt{\frac{61}{2}}\), i.e. all sides are equal.
But PR ≠ QS i.e. diagonals are not of equal measure.
Hence, the given figure is a rhombus.