ABCD is a rectangle formed by the points A (–1, –1), B(–1, 6), C (3, 6) and D (3, –1). P, Q, R and S are midpoints of sides AB, BC, CD and DA respectively. Show that diagonals of the quadrilateral PQRS bisect each other.

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Solution

Q.
ABCD is a rectangle formed by the points A (–1, –1), B(–1, 6), C (3, 6) and D (3, –1). P, Q, R and S are midpoints of sides AB, BC, CD and DA respectively. Show that diagonals of the quadrilateral PQRS bisect each other.

Ans.
Given ABCD is a rectangle and P, Q, R and S are mid-points of sides AB, BC, CD and DA.

coordinates of $P = (\frac{- 1 - 1}{2}, \frac{- 1 + 6}{2}) = (- 1, \frac{5}{2})$

coordinates of $Q = (\frac{- 1 + 3}{2}, \frac{6 + 6}{2}) = (1, 6)$

coordinates of $R = (\frac{3 + 3}{2}, \frac{6 - 1}{2}) = (3, \frac{5}{2})$

coordinates of $S = (\frac{3 - 1}{2}, \frac{- 1 - 1}{2}) = (1, - 1)$

Now, we shall find the mid points of PR & SQ.

Mid points of P & R which is point $O = (\frac{- 1 + 3}{2}, \frac{\frac{5}{2} + \frac{5}{2}}{2}) = (1, \frac{5}{2})$

Similarly, the midpoint of S and $Q = (\frac{1 + 1}{2}, \frac{6 - 1}{2}) = (1, \frac{5}{2})$

Since, the midpoints of PR & QS both have the same coordinate $(1, \frac{5}{2})$ .

Hence, diagonals PR and SQ bisect to each other.

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