Show that the points A(3, 0), B(4, 5), C(−1, 4) and D(−2, −1) are the vertices of a rhombus.

Solution:

The given points are A(3, 0), B(4, 5), C(− 1, 4) and D(− 2, − 1).

$A B=\sqrt{(3-4)^{2}+(0-5)^{2}}=\sqrt{(-1)^{2}+(-5)^{2}}$

$=\sqrt{1+25}=\sqrt{26}$

$B C=\sqrt{(4+1)^{2}+(5-4)^{2}}=\sqrt{(5)^{2}+(1)^{2}}$

$=\sqrt{25+1}=\sqrt{26}$

$C D=\sqrt{(-1+2)^{2}+(4+1)^{2}}=\sqrt{(1)^{2}+(5)^{2}}$

$=\sqrt{1+25}=\sqrt{26}$

$A D=\sqrt{(3+2)^{2}+(0+1)^{2}}=\sqrt{(5)^{2}+(1)^{2}}$

$=\sqrt{25+1}=\sqrt{26}$

$A C=\sqrt{(3+1)^{2}+(0-4)^{2}}=\sqrt{(4)^{2}+(-4)^{2}}$

$=\sqrt{16+16}=4 \sqrt{2}$

$B D=\sqrt{(4+2)^{2}+(5+1)^{2}}=\sqrt{(6)^{2}+(6)^{2}}$

$=\sqrt{36+36}=6 \sqrt{2}$

$\because A B=B C=C D=A D=6 \sqrt{2}$ and $A C \neq B D$

Therefore, the given points are the vertices of a rhombus.

Area $(=A B C D)=\frac{1}{2} \times A C \times B D$

$=\frac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2}=24$ sq. units

Hence, the area of the rhombus is 24 sq. units.