Question

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

Answer 1

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

${\displaystyle AB=\sqrt{{(6-8)}^2+{(1-2)}^2}=\sqrt{{(-2)}^2+{(-1)}^2}}$

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

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

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

${\displaystyle CD=\sqrt{{(9-7)}^2+{(4-3)}^2}=\sqrt{{\left(2\right)}^2+{\left(1\right)}^2}}$

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

${\displaystyle AD=\sqrt{{(7-6)}^2+{(3-1)}^2}=\sqrt{{\left(1\right)}^2+{\left(2\right)}^2}}$

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

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

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

${\displaystyle =BD=\sqrt{{(8-7)}^2+{(2-3)}^2}=\sqrt{{\left(1\right)}^2+{(-1)}^2}}$

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

${\displaystyle \because AB=BC=CD=AD=\sqrt{5}\text{and}AC\ne BD}$

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

Area${\displaystyle (\Delta ABCD)=\frac{1}{2}\times AC\times BD}$

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

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