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.

Answer 1

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

$PQ=\sqrt{{(3-2)}^2+{(4+1)}^2}=\sqrt{1^2+5^2}=\sqrt{26}units$

$QR=\sqrt{{(-2-3)}^2+{(3-4)}^2}=\sqrt{25+1}=\sqrt{26}units$

$RS=\sqrt{{(-3+2)}^2+{(-2-3)}^2}=\sqrt{1+25}=\sqrt{26}units$
$SP=\sqrt{{(-3-2)}^2+{(-2-3)}^2}=\sqrt{26}units$
$PR=\sqrt{{(-2-2)}^2+{(3+1)}^2}=\sqrt{16+16}=4\sqrt{2}units$
and, $QS=\sqrt{{(-3-3)}^2+{(-2-4)}^2}=\sqrt{36+36}=6\sqrt{2}units$
$\therefore PQ=QR=RS=SP=\sqrt{26}units$
and, $PR\ne QS$
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=$\frac{1}{2}\times \left(Productoflengthsofdiagonals\right)$

$\Rightarrow AreaofrhombusPQRS=\frac{1}{2}\times (PR\times QS)$
$\Rightarrow AreaofrhombusPQRS=(\frac{1}{2}\times 4\sqrt{2}\times 6\sqrt{2})sq.units=24sq.units$