Question

Find the equation for the line is perpendicular to 9x+3y=15 (-3,3)

Answer 1

well, let's keep in mind that perpendicular lines have negative reciprocal slopes, so hmmm wait a second, what's the slope of the equation above anyway?

$\bf 9x+3y=15\implies 3y=-9x+15\implies y=\cfrac{-9x+15}{3}\implies y=\cfrac{-9x}{3}+\cfrac{15}{3} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-3}x+5\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-3\implies -\cfrac{3}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{1}{3}}\qquad \stackrel{negative~reciprocal}{\cfrac{1}{3}}}$

so then, we're really looking for the equation of a line whose slope is 1/3 and runs through (-3,3)

$\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{3})~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{\cfrac{1}{3}}[x-\stackrel{x_1}{(-3)}] \\\\\\ y-3=\cfrac{1}{3}(x+3)\implies y-3=\cfrac{1}{3}+1\implies y=\cfrac{1}{3}x+4$

Answer 2

Answer:

-x + 3y = 12.

Step-by-step explanation:

OK. First we need to find the slope of 9x + 3y = 15.

3y = -9x + 15

y = -3x + 5

So it's slope = -3.

Now the slope of the line perpendicular to it is  - 1 / -3 = 1/3.

Now we use the general form point-slope to find the required equation:

y - y1 = m(x - x1)

Here x1 = - 3,  y1 = 3 and m = 1/3:

y - 3 = 1/3(x - -3)

y - 3 =  1`/3(x + 3)

Multiply through by 3:

3y - 9 = x + 3

-x + 3y = 12.