Question

Write the equation of a line that is perpendicular to y=-2/7+9 and that passes through the point (4,-6)

Answer 1

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

`y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{7}}x+9\qquad \impliedby \qquad \begin{array}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} \\\\[-0.35em] ~\dotfill`

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{ \cfrac{-2}{7}} ~\hfill \stackrel{reciprocal}{\cfrac{7}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{7}{-2} \implies \cfrac{7}{ 2 }}}`

so we're really looking for the equation of a line whose slope is 7/2 and it passes through (4 , -6)

`(\stackrel{x_1}{4}~,~\stackrel{y_1}{-6})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{7}{2} \\\\\\ \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}{(-6)}=\stackrel{m}{ \cfrac{7}{2}}(x-\stackrel{x_1}{4}) \implies y +6= \cfrac{7}{2} (x -4) \\\\\\ y+6=\cfrac{7}{2}x-14\implies {\Large \begin{array}{llll} y=\cfrac{7}{2}x-20 \end{array}}`