Question

line s passes through the points (2,7) and (6,1) line t passes through the point (5,2) and is perpendicular to line s what is the equation for line t

Answer 1

let's keep in mind that perpendicular lines have negative reciprocal slopes, hmmm what is the slope of line S anyway?

$\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-7}{6-2}\implies \cfrac{-6}{4}\implies -\cfrac{3}{2} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{3}{2}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{2}{3}}\qquad \stackrel{negative~reciprocal}{\cfrac{2}{3}}}$

so, we're really looking for the equation of a line whose slope is 2/3 and runs through (5,2)

$\bf (\stackrel{x_1}{5}~,~\stackrel{y_1}{2})~\hspace{10em} slope = m\implies \cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=\cfrac{2}{3}(x-5)\implies y-2=\cfrac{2}{3}x-\cfrac{10}{3} \\\\\\ y=\cfrac{2}{3}x-\cfrac{10}{3}+2\implies y=\cfrac{2}{3}x-\cfrac{4}{3}$