Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above
[tex]5x-y=-7\implies 5x-y+7=0 \\\\\\ \stackrel{\stackrel{m}{\downarrow }}{5} x+7=y\qquad \impliedby \begin{array}{|c|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}[/tex]
so a line perpendicular to that one above will have a slope of
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{5\implies \cfrac{5}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{5}}}[/tex]
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
