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Write an equation of the line passing through the point (2, 10) that is perpendicular to the line 2z+y=17

An equation of the line is y= blank x+blank

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


2x+y=17\implies y=\stackrel{\stackrel{m}{\downarrow }}{-2} x+17\qquad \impliedby \begin{array}c \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}{-2\implies \cfrac{-2}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-2}\implies \cfrac{1}{2}}}

so we're really looking for the equation of a line whose slope is 1/2 and that is passes through (2 , 10)


(\stackrel{x_1}{2}~,~\stackrel{y_1}{10})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{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}{10}=\stackrel{m}{ \cfrac{1}{2}}(x-\stackrel{x_1}{2}) \\\\\\ y-10=\cfrac{1}{2}x-1\implies {\Large \begin{array}{llll} y=\cfrac{1}{2}x+9 \end{array}}

User Amnesia
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