Respuesta :
well, first off let's check what is the slope of y = 2/3x+1.... .hmmmmm, notice [tex]\bf y=\stackrel{slope}{\cfrac{2}{3}}x+1[/tex] , since the equation is already in slope-intercept form, we can see it has a slope of 2/3.
well then, a line perpendicular to that one, will have a negative reciprocal to it.
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{2}{3}\\\\ slope=\cfrac{2}{{{ 3}}}\qquad negative\implies -\cfrac{2}{{{ 3}}}\qquad reciprocal\implies - \cfrac{{{ 3}}}{2}[/tex]
so, we're really looking for the equation of a line whose slope is -3/2 and goes through 12, -6.
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 12}}\quad ,&{{ -6}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies -\cfrac{3}{2} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-6)=-\cfrac{3}{2}(x-12) \\\\\\ y+6=-\cfrac{3}{2}x+18\implies y=-\cfrac{3}{2}x+12[/tex]
well then, a line perpendicular to that one, will have a negative reciprocal to it.
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{2}{3}\\\\ slope=\cfrac{2}{{{ 3}}}\qquad negative\implies -\cfrac{2}{{{ 3}}}\qquad reciprocal\implies - \cfrac{{{ 3}}}{2}[/tex]
so, we're really looking for the equation of a line whose slope is -3/2 and goes through 12, -6.
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 12}}\quad ,&{{ -6}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies -\cfrac{3}{2} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-6)=-\cfrac{3}{2}(x-12) \\\\\\ y+6=-\cfrac{3}{2}x+18\implies y=-\cfrac{3}{2}x+12[/tex]
