Question

is the line through points p(-8,-10) and q(-5,-12) perpendicular to the line through points r(9,-6) and s(17,-5)

Answer 1

Explanation:

No they are not perpendicular.

Perpendicular lines have slopes that are negative reciprocals of each other.

PQ slope = -12 - -10/-5 --8 = -2/3

RS slope = -5 - -6/17 -9 = 1/8

Slope are not negative reciprocals - the lines are not perpendicular.

Slope formula is m = y2 - y1/x2 - x1

Use this to determine slope.

For example if the the slope of RS was 3/2 - the lines would be perpendicular.

Answer 2

we dunno, hmmm let's check for the slope for PQ

`P(\stackrel{x_1}{-8}~,~\stackrel{y_1}{-10})\qquad Q(\stackrel{x_2}{-5}~,~\stackrel{y_2}{-12}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-12}-\stackrel{y1}{(-10)}}}{\underset{\textit{\large run}} {\underset{x_2}{-5}-\underset{x_1}{(-8)}}} \implies \cfrac{-12 +10}{-5 +8} \implies \cfrac{ -2 }{ 3 } \implies - \cfrac{2 }{ 3 }`

keeping in mind that perpendicular lines have negative reciprocal slopes, then if both are truly perpendicular, then line RS will have a slope of

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{ \cfrac{-2}{3}} ~\hfill \stackrel{reciprocal}{\cfrac{3}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{-2} \implies \cfrac{3}{ 2 }}}`

let's see if that's true

`R(\stackrel{x_1}{9}~,~\stackrel{y_1}{-6})\qquad S(\stackrel{x_2}{17}~,~\stackrel{y_2}{-5}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-5}-\stackrel{y1}{(-6)}}}{\underset{\textit{\large run}} {\underset{x_2}{17}-\underset{x_1}{9}}} \implies \cfrac{-5 +6}{8} \implies \cfrac{ 1 }{ 8 } ~~ \bigotimes ~~ \textit{not perpendicular}`