Question

Both equations I need, but I can graph it

Answer 1

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

`3x-5y=6\implies 3x=6+5y\implies 3x-6=5y\implies \cfrac{3x-6}{5}=y \\\\\\ \stackrel{ \stackrel{m}{\downarrow } }{\cfrac{3}{5}x}-\cfrac{6}{5}=y\qquad \impliedby \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{3}{5}} ~\hfill \stackrel{reciprocal}{\cfrac{5}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{5}{3} }}`

so we are really looking for the equation of a line whose slope is -5/3 and it passes through (-8 , 0)

`(\stackrel{x_1}{-8}~,~\stackrel{y_1}{0})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{5}{3} \\\\\\ \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}{0}=\stackrel{m}{-\cfrac{5}{3}}(x-\stackrel{x_1}{(-8)}) \\\\\\ y -0 = -\cfrac{5}{3} ( x +8) \implies {\Large \begin{array}{llll} y=-\cfrac{5}{3}x-\cfrac{40}{3} \end{array}}`

Answer 2

5x +3y = -40

Explanation:

You want an equation for the line through P(-8, 0) perpendicular to 3x -5y = 6.

Perpendicular line

The line perpendicular to ax +by = c can be written as ...

-bx +ay = c'

where c' is chosen so that the given point satisfies the equation.

For a=3, b=-5, this is ...

5x +3y = c'

For (x, y) = (-8, 0), the constant is ...

5(-8) +3(0) = c' = -40

The equation of the perpendicular line is 5x +3y = -40.

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Additional comment

The perpendicular line can be either of ...

-bx +ay = c'

bx -ay = c'

As a rule, we like the leading coefficient to be positive, so we choose the form of the equation to make that true. If 'a' and 'b' have a common factor, it usually works well to divide both coefficients by that factor, so they end up mutually prime.

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