Check the picture below.
[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\\\\ L(\stackrel{x_1}{-5}~,~\stackrel{y_1}{-7})\qquad M(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5}) ~\hfill LM=\sqrt{[ -2- -5]^2 + [ -5- -7]^2} \\\\\\ ~\hfill \boxed{LM=\sqrt{13}} \\\\\\ M(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-5})\qquad N(\stackrel{x_2}{-6}~,~\stackrel{y_2}{-4}) ~\hfill MN=\sqrt{[ -6- -2]^2 + [ -4- -5]^2} \\\\\\ ~\hfill \boxed{MN=\sqrt{17}}[/tex]
[tex]N(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-4})\qquad L(\stackrel{x_2}{-5}~,~\stackrel{y_2}{-7}) ~\hfill NL=\sqrt{[ -5- -6]^2 + [ -7- -4]^2} \\\\\\ ~\hfill \boxed{NL=\sqrt{10}}[/tex]
[tex]\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a = \sqrt{13}\\ b = \sqrt{17}\\ c = \sqrt{10}\\ s=\frac{\sqrt{13}+\sqrt{17}+\sqrt{10}}{2}\\ \qquad \approx 5.45 \end{cases} \\\\\\ A=\sqrt{5.45(5.45-\sqrt{13})(5.45-\sqrt{17})(5.45-\sqrt{10})}\implies \LARGE\textit{A=5.5}[/tex]
