Find the slope of any line perpendicular to the line passing through (−21,2) and (−32,5)

Francis Oliver
2022-10-18
Answered

Find the slope of any line perpendicular to the line passing through (−21,2) and (−32,5)

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Travis Sellers

Answered 2022-10-19
Author has **18** answers

First we need to find the slope of the line passing through the points: (−21,2) and (−32,5), the slope m between the points:

$({x}_{1},{y}_{1})\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}({x}_{2},{y}_{2})$ is given by:

$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$, so in this case:

$m=\frac{5-2}{-32-(-21)}$, simplifying we get:

$m=\frac{3}{-32+21}=\frac{3}{-11}=-\frac{3}{11}$

Now the perpendicular lines have slopes that are negative reciprocals, so if $m}_{1}\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}{m}_{2$ are the slopes of the two perpendicular lines then:

$m}_{2}=-\frac{1}{{m}_{1}$, therefore in this case:

$m}_{2}=-\frac{1}{-\frac{3}{11}}=\frac{11}{3$

$({x}_{1},{y}_{1})\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}({x}_{2},{y}_{2})$ is given by:

$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$, so in this case:

$m=\frac{5-2}{-32-(-21)}$, simplifying we get:

$m=\frac{3}{-32+21}=\frac{3}{-11}=-\frac{3}{11}$

Now the perpendicular lines have slopes that are negative reciprocals, so if $m}_{1}\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}{m}_{2$ are the slopes of the two perpendicular lines then:

$m}_{2}=-\frac{1}{{m}_{1}$, therefore in this case:

$m}_{2}=-\frac{1}{-\frac{3}{11}}=\frac{11}{3$

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