Question

Place (dlick and drag) one option from each of the lists below Into Its corresponding box to

create an equation of the line that passes through the polnt (1, "10) and is perpandicular to y=-1/3 x +5

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 }}{-\cfrac{1}{3}}x+5\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}{\cfrac{-1}{3}} ~\hfill \stackrel{reciprocal}{\cfrac{3}{-1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{-1}\implies 3}}`

so we're really looking for the equation of a line whose slope is 3 and passes through (1 , 10)

`(\stackrel{x_1}{1}~,~\stackrel{y_1}{10})\qquad \qquad \stackrel{slope}{m}\implies 3 \\\\\\ \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}{3}(x-\stackrel{x_1}{1}) \\\\\\ y-10=3x-3\implies y=3x+7`