Find the equation perpendicular to the line y=−3x+5 that passes through point (2, 6). Show Work

so hmmm, let's see, the equation above, is already in slope-intercept form, thus   [tex]\bf y=\stackrel{slope}{-3}x+5[/tex].

so, a perpendicular line to that, will have a negative reciprocal slope, that is, if the slope for that one is -3, then

[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope }-3\implies \cfrac{-3}{1}\\\\ slope=\cfrac{-3}{{{ 1}}}\qquad negative\implies +\cfrac{3}{{{1}}}\qquad reciprocal\implies +\cfrac{{{ 1}}}{3}[/tex]

so then, what is the equation of a line whose slope is 1/3 and goes through 2,6?

[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 2}}\quad ,&{{ 6}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{1}{3} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-6=\cfrac{1}{3}(x-2)\implies y-6=\cfrac{1}{3}x-\cfrac{2}{3} \\\\\\ y=\cfrac{1}{3}x-\cfrac{2}{3}+6\implies y=\cfrac{1}{3}x+\cfrac{16}{3}[/tex]