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I NEED HELP ON THIS PROBLEM!!! PLEASE


Find an equation for a line passing through (10,10) that is perpendicular to the line with equation 5x + 8y = -9. Write your equation in either point-slope form or slope-intercept form


Sagot :

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

[tex]5x+8y=-9\implies 8y=-5x-9\implies y=\cfrac{-5x-9}{8} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{5}{8}} x-9\qquad \impliedby \begin{array}{|c|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}[/tex]

well then, so since this equation has that slope therefore

[tex]\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{-5}{8}} ~\hfill \stackrel{reciprocal}{\cfrac{8}{-5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{8}{-5}\implies \cfrac{8}{5}}}[/tex]

so we're really looking for the equation of a line whose slope is 8/5 and runs through (10,10)

[tex](\stackrel{x_1}{10}~,~\stackrel{y_1}{10}) ~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{8}{5} \\\\\\ \begin{array}{|c|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}{\cfrac{8}{5}}(x-\stackrel{x_1}{10})[/tex]