Question

Line l contains the points (-1,-5) and (6,4). Line is perpendicular to line l and contains the point (9,3). what is the equation of line ?

Answer 1

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the line L above

`\stackrel{\textit{slope for line L}}{(\stackrel{x_1}{-1}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{4})} ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{4}-\stackrel{y1}{(-5)}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{(-1)}}} \implies \cfrac{4 +5}{6 +1}\implies \cfrac{9}{7} \\\\[-0.35em] ~\dotfill`

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{9}{7}} ~\hfill \stackrel{reciprocal}{\cfrac{7}{9}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{7}{9}}}`

so we're really looking for the equation of a line whose slope is -7/9 and that it passes through (9 , 3)

`(\stackrel{x_1}{9}~,~\stackrel{y_1}{3})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{7}{9} \\\\\\ \begin{array} \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}{3}=\stackrel{m}{- \cfrac{7}{9}}(x-\stackrel{x_1}{9}) \\\\\\ y-3=- \cfrac{7}{9}x+7\implies {\Large \begin{array}{llll} y=- \cfrac{7}{9}x+10 \end{array}}`