Show that the points A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a rhombus.

Solution:

The given points are A(6, 1), B(8, 2), C(9, 4) and D(7, 3).

$A B=\sqrt{(6-8)^{2}+(1-2)^{2}}=\sqrt{(-2)^{2}+(-1)^{2}}$

$=\sqrt{4+1}=\sqrt{5}$

$B C=\sqrt{(8-9)^{2}+(2-4)^{2}}=\sqrt{(-1)^{2}+(-2)^{2}}$

$=\sqrt{1+4}=\sqrt{5}$

$C D=\sqrt{(9-7)^{2}+(4-3)^{2}}=\sqrt{(2)^{2}+(1)^{2}}$

$=\sqrt{4+1}=\sqrt{5}$

$A D=\sqrt{(7-6)^{2}+(3-1)^{2}}=\sqrt{(1)^{2}+(2)^{2}}$

$=\sqrt{1+4}=\sqrt{5}$

$A C=\sqrt{(6-9)^{2}+(1-4)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}$

$=\sqrt{9+9}=3 \sqrt{2}$

$B D=\sqrt{(8-7)^{2}+(2-3)^{2}}=\sqrt{(1)^{2}+(-1)^{2}}$

$=\sqrt{1+1}=\sqrt{2}$

$\because A B=B C=C D=A D=\sqrt{5}$ and $A C \neq B D$

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

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

$=\frac{1}{2} \times 3 \sqrt{2} \times \sqrt{2}=3$ sq. units

Hence, the area of the rhombus is 3 sq. units.