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# 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, Q, R and S are the midpoints of AB, BC, CD and DA respectively

So, coordinate of P
= ( $$\frac{-1-1}{2} , \frac{-1+4}{2}$$ )
= ( $$-1 , \frac{3}{2}$$ )

Similarly, coordinate of Q
= ( $$\frac{5-1}{2} , \frac{4+4}{2}$$ )
= ( 2,4)

R = ( $$\frac{5+5}{2} , \frac{4-1}{2}$$ ) = ( $$5 , \frac{3}{2}$$ )

S = ( $$\frac{5-1}{2} , \frac{-1-1}{2}$$ ) = ( 2,-1)

Now, PQ = $$\sqrt{(-1-2)^2 + ( \frac{3}{2} - 4)^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$

SP = $$\sqrt{(2+1)^2 + (-1- \frac{3}{2})^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$

QR = $$\sqrt{(2-5)^2 + (4- \frac{3}{2})^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$

RS = $$\sqrt{(5-2)^2 + ( \frac{3}{2} +1)^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$

PR = $$\sqrt{(-1-5)^2 + ( \frac{3}{2} - \frac{3}{2})^2 } \$$
$$= \ 6$$

QS = $$\sqrt{(2-2)^2 + (4+1)^2} \$$
$$= \ 5$$

Clearly, PQ = QR = RS = SP

But, $$PR \ne QS$$

Hence the lengths of all the sides are equal, while the lengths of the diagonals are different.

$$\therefore$$ The PQRS is a rhombus.