Question

Write an equation

intercept form for a line
in slope-
perpendicular to y = -2x + 6
containing (3, 2).

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{m}{\downarrow }}{-2}x+6\qquad \impliedby \qquad \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 it passes through (3 , 2)

`(\stackrel{x_1}{3}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{2} \\\\\\ \begin{array} \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}{2}=\stackrel{m}{ \cfrac{1}{2}}(x-\stackrel{x_1}{3}) \\\\\\ y-2=\cfrac{1}{2}x-\cfrac{3}{2}\implies y=\cfrac{1}{2}x-\cfrac{3}{2}+2\implies {\Large \begin{array}{llll} y=\cfrac{1}{2}x+\cfrac{1}{2} \end{array}}`