Respuesta :
let's take a peek at 7x + 4 for a second, [tex]\bf y = \stackrel{slope}{7}x\stackrel{y-intercept}{+4}[/tex]
well well, it has a slope of 7.
now, a line perpendicular to that one, will have a negative reciprocal slope,
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad 7\implies \cfrac{7}{1}\\\\ negative\implies -\cfrac{7}{{{ 1}}}\qquad reciprocal\implies - \cfrac{{{ 1}}}{7}[/tex]
so, we're really looking for the equation of a line whose slope is -1/7 and runs through -4, -2
[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1\\ % (a,b) &&(~{{ -4}} &,&{{ -2}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies -\cfrac{1}{7} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-2)=-\cfrac{1}{7}[x-(-4)] \\\\\\ y+2=-\cfrac{1}{7}(x+4)\implies y+2=-\cfrac{1}{7}x-\cfrac{4}{7} \\\\\\ y=-\cfrac{1}{7}x-\cfrac{4}{7}-2\implies y=-\cfrac{1}{7}x-\cfrac{18}{7}[/tex]
well well, it has a slope of 7.
now, a line perpendicular to that one, will have a negative reciprocal slope,
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad 7\implies \cfrac{7}{1}\\\\ negative\implies -\cfrac{7}{{{ 1}}}\qquad reciprocal\implies - \cfrac{{{ 1}}}{7}[/tex]
so, we're really looking for the equation of a line whose slope is -1/7 and runs through -4, -2
[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1\\ % (a,b) &&(~{{ -4}} &,&{{ -2}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies -\cfrac{1}{7} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-2)=-\cfrac{1}{7}[x-(-4)] \\\\\\ y+2=-\cfrac{1}{7}(x+4)\implies y+2=-\cfrac{1}{7}x-\cfrac{4}{7} \\\\\\ y=-\cfrac{1}{7}x-\cfrac{4}{7}-2\implies y=-\cfrac{1}{7}x-\cfrac{18}{7}[/tex]
