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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 √(x1−x2)2+(y1−y2)2 PQ =√(3−2)2+(4+1)2= √12+52 = √26 units. QR = √(−2−3)2+(3−4)2= √25+1 = √26 units. RS = √(−3+2)2+(−2−3)2 = √1+25 = √26 units. SP = √(−3−2)2+(−2+1)2 = √26 units. PR = √(−2−2)2+(3+1)2 = √16+16 = 4 √2 units and, QS = √(−3−3)2+(−2−4)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 ×6√2)sq. units = 24 sq. units

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