Find the slope of any line perpendicular to the line passing through (−2,17) and (2,8)

Raiden Barr
2022-10-15
Answered

Find the slope of any line perpendicular to the line passing through (−2,17) and (2,8)

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Martha Dickson

Answered 2022-10-16
Author has **20** answers

If you have 2 points you can find the slope of the line joining them from the formula:

$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

$m=\frac{17-8}{-2-2}=\frac{9}{-4}$

Perpendicular lines have the following properties:

They intersect at 90°

Their slopes are exactly opposite ...

Where one is steep, the other is gentle.

If one is positive, the other is negative.

One slope is the negative reciprocal of the other.

If $m}_{1}=\frac{a}{b},\phantom{\rule{1ex}{0ex}}\text{then}\phantom{\rule{1ex}{0ex}}{m}_{2}=-\frac{b}{a$

The product of their slopes is -1

${m}_{1}\times {m}_{2}=-1$

So in this case:

$m}_{1}=-\frac{9}{4}\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\to \phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{m}_{2}=\frac{4}{9$

$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

$m=\frac{17-8}{-2-2}=\frac{9}{-4}$

Perpendicular lines have the following properties:

They intersect at 90°

Their slopes are exactly opposite ...

Where one is steep, the other is gentle.

If one is positive, the other is negative.

One slope is the negative reciprocal of the other.

If $m}_{1}=\frac{a}{b},\phantom{\rule{1ex}{0ex}}\text{then}\phantom{\rule{1ex}{0ex}}{m}_{2}=-\frac{b}{a$

The product of their slopes is -1

${m}_{1}\times {m}_{2}=-1$

So in this case:

$m}_{1}=-\frac{9}{4}\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\to \phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{m}_{2}=\frac{4}{9$

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