bearing in mind that parallel lines have exactly the same slope, hmmm what's the slope of that equation above anyway?
[tex]\bf -x+y=5\implies y=x+5\implies y = \stackrel{\stackrel{m}{\downarrow }}{1}x+5\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]
so we're really looking for the equation of a line whose slope is 1 and runs through (2,-5)
[tex]\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{-5})~\hspace{10em} \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}{(-5)}=\stackrel{m}{1}(x-\stackrel{x_1}{2}) \\\\\\ y+5=x-2\implies y=x-7[/tex]
