Question

What is perpendicular to y=-(1/3)x+5 but goes though the point (1,-10) in slope-intersect form

Answer 1

use M=(Y-Y)/(X-X)format and see what you get

Explanation:

so it should be

-10-3/1-1

Im not sure how to explain this without having the proper set up

(2,2) = (x,y)

(1,1) = (x,y)

so if we set this up right

-13/0....

Answer 2

bearing in mind that perpendicular lines have negative reciprocal slopes hmmmmm wait a second, what's the slope of that line above anyway?

`\bf y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{1}{3}} x+5\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}`

therefore any perpendicular line to that

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

so, we're really looking for the equation of a line whose slope is 3 and runs through (1, -10)

`\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{-10})~\hspace{10em} slope = m\implies 3 \\\\\\ \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-(-10)=3(x-1) \\\\\\ y+10=3x-3\implies y=3x-13`