Christopher Saunders

## Answered question

2022-10-15

I want to find the equation of line perpendicular to $y=-\frac{5}{3}x-8$ and passing through (−9,−3)

### Answer & Explanation

flasheadita237m

Beginner2022-10-16Added 17 answers

The two lines are perpendicular to each other, if the product of their slopes is −1. Hence if slope of one line is given as $\frac{a}{b}$, slope of line perpendicular to it is $-\frac{b}{a}$.
As equation of one line is given in slope-intercept form as $y=-\frac{5}{3}x-8$, its slope is $-\frac{5}{3}$.
Hence slope of line perpendicular to it is $\frac{3}{5}$.
Now equation of line passing through $\left({x}_{1},{y}_{1}\right)$ and having slope m is $\left(y-{y}_{1}\right)=m\left(x-{x}_{1}\right)$
Hence, equation of a line passing through (−9,−3) and having a slope $\frac{3}{5}$ is
$\left(y-\left(-3\right)\right)=\frac{3}{5}\left(x-\left(-9\right)\right)$
or 5(y+3)=3(x+9)
or 5y+15=3x+27
or 3x−5y+12=0

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