Question

Open this doc 2 (less blurry one!)

Answer 1

m = 1/3

m⊥ = -3

x1 = -9

y1 = 13

point slope: y-13 = -3(x+9)

slope intercept: y= -3x-14

Explanation:

Equation given is in y=mx+b form, so I think m=1/3 (or whatever is right before the x cause it is little hard to see)

To find a perpendicular slope, you use the negative reciprocal of the given slope, so the perpendicular slope would be m=-3

x1 and y1 would be the given coordinates so -9 and 13 respectively

Finally plug in all this info into point slope form equation (y-y1 = m(x-x1)

Point slope would be: y-13 = -3(x+9)

Slope intercept would be: y= -3x-14

Hope this helps!

Answer 2

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{1}{3}}x+16\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{1}{3}} ~\hfill \stackrel{reciprocal}{\cfrac{3}{1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{1} \implies -3}}`

so we're really looking for the equation of a line whose slope is -3 and it passes through (-9 , 13)

`(\stackrel{x_1}{-9}~,~\stackrel{y_1}{13})\hspace{10em} \stackrel{slope}{m} ~=~ - 3 \\\\\\ \begin{array} \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}{13}=\stackrel{m}{- 3}(x-\stackrel{x_1}{(-9)}) \implies y -13 = - 3 ( x +9) \\\\\\ y-13-3x-27\implies {\Large \begin{array}{llll} y=-3x-14 \end{array}}`