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 }}{-1}x+2\qquad \impliedby \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]
now, we know that our line is perpendicular to that one, thus
[tex]\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-1\implies \cfrac{-1}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-1}\implies 1}}[/tex]
so we're really looking for the equation of a line that has a slope of 1 and runs through (0 , 10)
[tex](\stackrel{x_1}{0}~,~\stackrel{y_1}{10})\qquad \qquad \stackrel{slope}{m}\implies 1 \\\\\\ \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}{10}=\stackrel{m}{1}(x-\stackrel{x_1}{0}) \\\\\\ y-10=x\implies y=x+10[/tex]
