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What is the slope of a line perpendicular to the line whose equation x+y=3. Fully simplify your answer

Sagot :

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

[tex]x + y =3\implies y=-x+3\implies y=\stackrel{\stackrel{m}{\downarrow }}{-1}x+3 \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} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{-1\implies \cfrac{-1}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-1}\implies 1}}[/tex]

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