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 mid-points of sides AB, BC, CD and DA respectively. Show that diagonals of the quadrilateral PQRS bisect each other.

Ans: Let’s make a diagram for the given question:

ABCD is a rectangle formed by the points A (−1, −1), B (−1, 6), C (3, 6) and D (3, −1).

Let’s start fiding coordinates of points P, Q, R and S.

We know that the coordinates of a midpoint lying between (x1, y1) and (x2, y2) is given by:

M(x,y) = (\frac{X_1 + X_2}{2}, \frac{Y_1 + Y_2}{2})

Since P is the midpoint of AB, ∴ P = (\frac{(- 1) + (- 1)}{2}, \frac{(- 1) + 6}{2}) = (- 1, \frac{5}{2})

Similarly, Q is the midpoint of BC, ∴ P = (\frac{(- 1) + 3}{2}, \frac{6 + 6}{2}) = (1, 6)

Similarly, R is the midpoint of CD, ∴ P = (\frac{3 + 3}{2}, \frac{6 + (- 1)}{2}) = (3, \frac{5}{2})

Similarly, S is the midpoint of DA, ∴ P = (\frac{3 + (- 1)}{2}, \frac{(- 1) + (- 1)}{2}) = (1, – 1)

If Diagonals of PQRS bisect each other, O will be the midpoint of PR and QS both.

Let’s check if this relationship is verified:

If O is the midpoint of PR, then its coordinates are given by: (\frac{(-1) +3}{2}, \frac{\frac{5}{2} + \frac{5}{2}}{2}) = (1,\frac{5}{2})

If O is the midpoint of QS, then its coordinates are given by: (\frac{1 + 1}{2}, \frac{6 + (- 1)}{2}) = (1,\frac{5}{2})

Since both points coordinates match with each other, it confirms that O is midpoint of PR & QS

Therefore, diagonals of quadrilateral PQRS bisect each other.

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