Question

The equation of a line is y=-x-3. Find the equation of a perpendicular line that goes through the point (-4, 1)​

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{\textit{\small m}}{\downarrow }}{-1}x-3\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}{ -1 \implies \cfrac{-1}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-1} \implies 1}}`

so we are really looking for the equation of a line whose slope is 1 and it passes through (-4 , 1)

`(\stackrel{x_1}{-4}~,~\stackrel{y_1}{1})\hspace{10em} \stackrel{slope}{m} ~=~ 1 \\\\\\ \begin{array}c \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}{1}=\stackrel{m}{1}(x-\stackrel{x_1}{(-4)}) \implies y -1 = 1 ( x +4) \\\\\\ y -1 = x +4 \implies {\Large \begin{array}{llll} y = x +5 \end{array}}`