Question

write and equation for the line in point-slope form that passes through (2,-5) and is perpendicular to the line 2x-4y+8=0​

Answer 1

bearing in mind that perpendicular lines have negative reciprocal slopes, hmmm what's the slope of the equation above?

`\bf 2x-4y+8=0\implies -4y=-2x-8\implies y = \cfrac{-2x-8}{-4} \\\\\\ y = \cfrac{-2x}{4}-\cfrac{8}{-4}\implies y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{1}{2}}x+2\qquad \impliedby \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`

`\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{1}{2}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{2}{1}}\qquad \stackrel{negative~reciprocal}{\cfrac{2}{1}\implies 2}}`

so we're really looking for the equation of a line whose slope is 2 and runs through (2,-5),

`\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{-5})~\hspace{10em} \stackrel{slope}{m}\implies 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}{(-5)}=\stackrel{m}{2}(x-\stackrel{x_1}{2}) \\\\\\ y+5=2x-4\implies y=2x-9`