Question

What is the equation of a line that is perpendicular to the line y=2x+1 and passes through the point (4,6).

i need help asap! work shown to help me thru the steps. y=mx+b

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+1\qquad \impliedby \qquad \begin{array} \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}}}`

so we're really looking for the equation of a line with a slope of -1/2 and that it passes through (4 , 6)

`(\stackrel{x_1}{4}~,~\stackrel{y_1}{6})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{1}{2} \\\\\\ \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}{6}=\stackrel{m}{- \cfrac{1}{2}}(x-\stackrel{x_1}{4}) \\\\\\ y-6=- \cfrac{1}{2}x+2\implies {\Large \begin{array}{llll} y=- \cfrac{1}{2}x+8 \end{array}}`