What is the equation, in point-slope form, of the line that is perpendicular to the given line and passes through thepoint (2,5)?y + 5 = x + 2y-2 = x-5y-5=-(x - 2)y + 2 = -(x + 5)

keeping in mind that perpendicular lines have negative reciprocal slopes, hmmmm what's the slope of that line above anyway,[tex]\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{1}}}\implies \cfrac{2+1}{3}\implies 1 \\\\[-0.35em] ~\dotfill[/tex][tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\underline{1}\implies \cfrac{\underline{1}}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{\underline{1}}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{\underline{1}}\implies -1}}[/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]