Respuesta :
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above
[tex]y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{9}} x-7\qquad \impliedby \qquad \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]
so then we can say that
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{-\cfrac{2}{9}} ~\hfill \stackrel{reciprocal}{-\cfrac{9}{2}} ~\hfill \stackrel{negative~reciprocal}{-\left( -\cfrac{9}{2} \right)\implies \cfrac{9}{2}}}[/tex]
so we're really looking for the equation of a line whos slope is 9/2 and passes through (2 , 3)
[tex](\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{9}{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-\stackrel{y_1}{3}=\stackrel{m}{\cfrac{9}{2}}(x-\stackrel{x_1}{2}) \\\\\\ y-3=\cfrac{9}{2}x-9\implies \boxed{y=\cfrac{9}{2}x-6}[/tex]
now as far as the parallel line
keeping in mind that parallel lines have exactly the same slope, so we're really looking for the equation of a line whose slope is -2/9 and passes through (2 , 3)
[tex](\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{2}{9} \\\\\\ \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-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{2}{9}}(x-\stackrel{x_1}{2}) \\\\\\ y-3=-\cfrac{2}{9}x+\cfrac{4}{9}\implies y=-\cfrac{2}{9}x+\cfrac{4}{9}+3\implies \boxed{y=-\cfrac{2}{9}x+\cfrac{31}{9}}[/tex]
