Respuesta :
so the triangle has the vertices of (2, 18) (-2, -4), and (6,12), that gives us the endpoints for each line of
(2, 18) , (-2, -4)
(-2, -4) , (6,12)
(6,12) , (2, 18)
[tex]\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{18})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-4}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-4-18}{-2-2}\implies \cfrac{-22}{-4}\implies \cfrac{11}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-18=\cfrac{11}{2}(x-2) \\\\\\ y-18=\cfrac{11}{2}x-11 \implies \blacktriangleright y=\cfrac{11}{2}x+7 \blacktriangleleft[/tex]
[tex]\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-(-4)}{6-(-2)}\implies \cfrac{12+4}{6+2}\implies \cfrac{16}{8}\implies 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-4)=2[x-(-2)] \\\\\\ y+4=2(x+2)\implies y+4=2x+4\implies \blacktriangleright y=2x \blacktriangleleft[/tex]
[tex]\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{12})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{18}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{18-12}{2-6}\implies \cfrac{6}{-3}\implies -2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-12=-2(x-6) \\\\\\ y-12=-2x+12\implies \blacktriangleright y=-2x+24 \blacktriangleleft[/tex]
