Question

triangle lmn, with vertices L(-5,-7), M(-2,-5), and N(-6,-4), is drawn on the coordinate grid bellow. what is the area, in square units, of triangle LMN?

Answer 1

Check the picture below.

`~\hfill \stackrel{\textit{\large distance between 2 points}}{d = √(( 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=√([ -2- -5]^2 + [ -5- -7]^2) \\\\\\ ~\hfill \boxed{LM=√(13)} \\\\\\ M(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-5})\qquad N(\stackrel{x_2}{-6}~,~\stackrel{y_2}{-4}) ~\hfill MN=√([ -6- -2]^2 + [ -4- -5]^2) \\\\\\ ~\hfill \boxed{MN=√(17)}`

`N(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-4})\qquad L(\stackrel{x_2}{-5}~,~\stackrel{y_2}{-7}) ~\hfill NL=√([ -5- -6]^2 + [ -7- -4]^2) \\\\\\ ~\hfill \boxed{NL=√(10)}`

`\qquad \textit{Heron's area formula} \\\\ A=√(s(s-a)(s-b)(s-c))\qquad \begin{cases} s=(a+b+c)/(2)\\[-0.5em] \hrulefill\\ a = √(13)\\ b = √(17)\\ c = √(10)\\ s=(√(13)+√(17)+√(10))/(2)\\ \qquad \approx 5.45 \end{cases} \\\\\\ A=\sqrt{5.45(5.45-√(13))(5.45-√(17))(5.45-√(10))}\implies \LARGE\textit{A=5.5}`

Answer 2

53

Explanation:

but