How do you find the slope that is perpendicular to the line $- 5 x - 3 y + 8 = 0$ $-5x-3y+8=0$ ?

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Answer 1 · 2016-08-26T23:36:04

The slope of one line and the slope of a line perpendicular to the first line are negative inverses of each other.

If Line1 has slope $m = 3$ $m = 3$ , then perpendicular Line2 will have slope $m = - \frac{1}{3}$ $m = -1/3$

To find the slope of your line $- 5 x - 3 y + 8 = 0$ $−5x−3y+8=0$ , let's put the equation in slope-intercept form:

$y = - (\frac{5}{3}) x + \frac{8}{3}$ $y = - (5/3)x + 8/3$

So, your line has slope $m = - \frac{5}{3}$ $m =- 5/3$ .

The slope of a line that is perpendicular to your line above will be the negative inverse, that is:

$\frac{- 1}{- \frac{5}{3}}$ $(-1)/(-5/3)$ , which equals positive $\frac{3}{5}$ $3/5$ .

(It's the lines that are perpendicular to each other, not the slopes. The slopes are negative inverses.)

How do you find the slope that is perpendicular to the line -5x-3y+8=0−5x−3y+8=0-5x-3y+8=0?