We are given rectangle with vertices (6, −3) , (3, −6) , (−1, −2) , and (2, 1) .
[tex]\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}[/tex]
[tex]\mathrm{The\:distance\:between\:}\left(6,\:-3\right)\mathrm{\:and\:}\left(3,\:-6\right)\mathrm{\:is\:}[/tex]
[tex]=\sqrt{\left(3-6\right)^2+\left(-6-\left(-3\right)\right)^2}\[/tex]
Length of the rectangle is [tex]=3\sqrt{2}[/tex].
Width of the rectangle is
[tex]=\sqrt{\left(-1-3\right)^2+\left(-2-\left(-6\right)\right)^2}[/tex]
[tex]=4\sqrt{2}[/tex]
Area of the rectangle = length × width.
Plugging values of length and width in above formula, we get
Area = [tex]3\sqrt{2}\times 4\sqrt{2}\:[/tex]= 12(2) =24 square units.
