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
$P Q = \sqrt{{(3 - 2)}^{2} + {(4 + 1)}^{2}} = \sqrt{1^{2} + 5^{2}} = \sqrt{26} u n i t s$
$Q R = \sqrt{{(- 2 - 3)}^{2} + {(3 - 4)}^{2}} = \sqrt{25 + 1} = \sqrt{26} u n i t s$
$R S = \sqrt{{(- 3 + 2)}^{2} + {(- 2 - 3)}^{2}} = \sqrt{1 + 25} = \sqrt{26} u n i t s$
$S P = \sqrt{{(- 3 - 2)}^{2} + {(- 2 - 3)}^{2}} = \sqrt{26} u n i t s$
$P R = \sqrt{{(- 2 - 2)}^{2} + {(3 + 1)}^{2}} = \sqrt{16 + 16} = 4 \sqrt{2} u n i t s$
and, $Q S = \sqrt{{(- 3 - 3)}^{2} + {(- 2 - 4)}^{2}} = \sqrt{36 + 36} = 6 \sqrt{2} u n i t s$
$∴ P Q = Q R = R S = S P = \sqrt{26} u n i t s$
and, $P R \neq Q S$
This means that PQRS is quadrilateral whose sides are equal but diagonals are not equal.
Thus, PQRS is a rhombus but not a square.
.Now, Area of rhombus PQRS= $\frac{1}{2} \times (P r o d u c t o f l e n g t h s o f d i a g o n a l s)$
$\Rightarrow A r e a o f r h o m b u s P Q R S = \frac{1}{2} \times (P R \times Q S)$
$\Rightarrow A r e a o f r h o m b u s P Q R S = (\frac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2}) s q . u n i t s = 24 s q . u n i t s$

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