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Question

If P (2, – 1), Q(3, 4), R(–2, 3) and S(–3, –2) be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus.

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Solution

The given points are P(2, -1), Q(3, 4), R(-2, 3) and S(-3, -2).

We have,
Distance between the points is given by
(x1x2)2+(y1y2)2

PQ =(32)2+(4+1)2= 12+52 = 26 units.

QR = (23)2+(34)2= 25+1 = 26 units.

RS = (3+2)2+(23)2 = 1+25 = 26 units.

SP = (32)2+(2+1)2 = 26 units.

PR = (22)2+(3+1)2 = 16+16 = 4 2 units

and, QS = (33)2+(24)2

= 36+36 = 6 2 units

∴ PQ = QR = RS = SP = 26 units

and, PR ≠ QS

This means that PQRS is a quadrilateral whose sides are equal but diagonals are not equal.

Thus, PQRS is a rhombus but not a square.

Now, Area of rhombus PQRS = 12 ×(Product of lengths of diagonals)

⇒Area of rhombus PQRS = 12 × (PR × QS)

⇒Area of rhombus PQRS

=(12 × 4 2 ×62)sq. units = 24 sq. units


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