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,
Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$

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 = $\sqrt{{(-3-2)}^2+{(-2+1)}^2}$ = $\sqrt{26}$ 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

∴ PQ = QR = RS = SP = $\sqrt{26}$ 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 = $\frac{1}{2}$ ×(Product of lengths of diagonals)

⇒Area of rhombus PQRS = $\frac{1}{2}$ × (PR × QS)

⇒Area of rhombus PQRS

=($\frac{1}{2}$ × 4 $\sqrt{2}$ ×6$\sqrt{2}$)sq. units = 24 sq. units