Question

If anyone know how to do this comment

Answer 1

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

`-8x+3y=-6\implies 3y=8x-6\implies y=\cfrac{8x-6}{3} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{8}{3}}x-2\qquad \impliedby \begin{array}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} \\\\[-0.35em] ~\dotfill`

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{ \cfrac{8}{3}} ~\hfill \stackrel{reciprocal}{\cfrac{3}{8}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{8} }}`

so we're really looking for the equation of a line whose slope is -3/8 and it passes through (-16 , 8)

`(\stackrel{x_1}{-16}~,~\stackrel{y_1}{8})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{3}{8} \\\\\\ \begin{array}c \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}{8}=\stackrel{m}{- \cfrac{3}{8}}(x-\stackrel{x_1}{(-16)}) \implies y -8= -\cfrac{3}{8} (x +16) \\\\\\ y-8=-\cfrac{3}{8}x-6\implies {\Large \begin{array}{llll} y=-\cfrac{3}{8}x+2 \end{array}}`

Answer 2

`\displaystyle y=-(3)/(8)x+2`

Explanation:

Perpendicular lines have opposite reciprocal slopes. So, we must first determine our slope of the original equation by converting to slope-intercept form:

`-8x+3y=-6\\3y=8x-6\\y=(8)/(3)x-2`

Hence, because the slope of the original line is 8/3, then the slope of the perpendicular line is -3/8. We are not done, however, as we need to account for the point that the perpendicular line passes through by solving for the new y-intercept:

`y=-(3)/(8)x+b\\ \\8=-(3)/(8)(-16)+b\\\\8=6+b\\\\2=b`

Thus, the perpendicular line equation is
`y=-(3)/(8)x+2`. See below for a graph.