Question

IXL: S26 (ALGEBRA 1)

The equation of line k is y=9/8x-9. Line I is perpendicular to line k and passes through (-2,2) What is the equation of line I? write the equation in slope intercept form. ​

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{9}{8}}x-9\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}{ \cfrac{9}{8}} ~\hfill \stackrel{reciprocal}{\cfrac{8}{9}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{8}{9} }}`

so we're really looking for the equation of a line whose slope is -8/9 and it passes through (-2 , 2)

`(\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{8}{9} \\\\\\ \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}{2}=\stackrel{m}{- \cfrac{8}{9}}(x-\stackrel{x_1}{(-2)}) \implies y -2= -\cfrac{8}{9} (x +2) \\\\\\ y-2=-\cfrac{8}{9}x-\cfrac{16}{9}\implies y=-\cfrac{8}{9}x-\cfrac{16}{9}+2\implies {\Large \begin{array}{llll} y=-\cfrac{8}{9}x+\cfrac{2}{9} \end{array}}`