Respuesta :
now, perpendicular lines have negative reciprocal slopes, hmm wait a second, what is the slope of y = 13x + 2 anyway? [tex]\bf y=\stackrel{slope}{13}x+2[/tex]
well then
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{13\implies \cfrac{13}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{13}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{13}}}[/tex]
so we're really looking for the equation of a line whose slope is -1/13 and runs through -5,2.
[tex]\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{2})\qquad \qquad \qquad slope = m\implies -\cfrac{1}{13} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-2=-\cfrac{1}{13}[x-(-5)] \\\\\\ y-2=-\cfrac{1}{13}(x+5)\implies y-2=-\cfrac{1}{13}x-\cfrac{5}{13} \\\\\\ y=-\cfrac{1}{13}x-\cfrac{5}{13}+2\implies y=-\cfrac{1}{13}x+\cfrac{21}{13}[/tex]
well then
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{13\implies \cfrac{13}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{13}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{13}}}[/tex]
so we're really looking for the equation of a line whose slope is -1/13 and runs through -5,2.
[tex]\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{2})\qquad \qquad \qquad slope = m\implies -\cfrac{1}{13} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-2=-\cfrac{1}{13}[x-(-5)] \\\\\\ y-2=-\cfrac{1}{13}(x+5)\implies y-2=-\cfrac{1}{13}x-\cfrac{5}{13} \\\\\\ y=-\cfrac{1}{13}x-\cfrac{5}{13}+2\implies y=-\cfrac{1}{13}x+\cfrac{21}{13}[/tex]
