Question

What is an equation of the line that passes through the point (4, -3)and is perpendicular to the line 2x-5y=35

Answer 1

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

`2x-5y=35\implies 2x=35+5y\implies 2x-35=5y \\\\\\ \cfrac{2x-35}{5}=y\implies \stackrel{\stackrel{m}{\downarrow }}{\cfrac{2}{5}}x-7=y\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}{5}} ~\hfill \stackrel{reciprocal}{\cfrac{5}{2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{5}{2} }}`

so we're really looking for the equation of a line whose slope is -5/2 and it passes through (4 , -3)

`(\stackrel{x_1}{4}~,~\stackrel{y_1}{-3})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{5}{2} \\\\\\ \begin{array}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}{(-3)}=\stackrel{m}{-\cfrac{5}{2}}(x-\stackrel{x_1}{4}) \implies y +3 = -\cfrac{5}{2} ( x -4) \\\\\\ y+3=-\cfrac{5}{2}x+10\implies {\Large \begin{array}{llll} y=-\cfrac{5}{2}x+7 \end{array}}`