Lines m and p are perpendicular. if the slope of line m is -3 what is the slope of line p

1 answer:

ivann1987 [24]3 years ago

4 0

For two lines to be perpendicular, their slopes must be negative reciprocals of one another, mathematically:

m1*m2=-1, in this case our reference line has a slope of -3 so:

-3m=-1 divide both sides by -3

m=-1/-3

m=1/3

So the perpendicular line will have a slope of 1/3

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Please help!! easy but i don't understand <3

Sati [7]

in essence it's just asking on getting the equation of that line in more or less slope-intercept form, hmmm well, it's a line, so all we need is two points to get it, hmmm let's see let's use the starting point and terminal point, so we know it goes through the origin, (0,0) and also goes through (1, 0.7), let's change 0.7 to a fraction, so its 7/10, ok, let's use those two points from it.

$(\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{\frac{7}{10}}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{\frac{7}{10}}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{0}}}\implies \cfrac{~~ \frac{7}{10}~~}{1}\implies \cfrac{7}{10}$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{\cfrac{7}{10}}(x-\stackrel{x_1}{0})\implies y = \cfrac{7}{10}x$