well, we know it's a rectangle, so we know the opposite sides are perpendicular, so if we just get say the sides AD as its length and BA as its width, that'd give us its area, since it'll just be AD * BA.
[tex]~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{-1}~,~\stackrel{y_1}{4})\qquad D(\stackrel{x_2}{-3}~,~\stackrel{y_2}{-4}) ~\hfill AD=\sqrt{(~~ -3- (-1)~~)^2 + (~~ -4- 4~~)^2} \\\\\\ ~\hfill AD=\sqrt{( -2)^2 + ( -8)^2}\implies \boxed{AD=\sqrt{68}} \\\\\\ B(\stackrel{x_1}{3}~,~\stackrel{y_1}{3})\qquad A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{4}) ~\hfill BA=\sqrt{(~~ -1- 3~~)^2 + (~~ 4- 3~~)^2}[/tex]
[tex]~\hfill BA=\sqrt{( -4)^2 + (1)^2} \implies \boxed{BA=3} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\LARGE Area}}{(\sqrt{68})(3)\implies (\sqrt{2^2\cdot 17})(3)}\implies (2\sqrt{17})( 3)\implies \LARGE \begin{array}{llll} 6\sqrt{17} \end{array}[/tex]
