Question

Find the equation of line perpendicular to `y=-5/3x-8` and passing through `(-9,-3)`?

Answer 1

`3x-5y+12=0`

Explanation:

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 `a/b`, slope of line perpendicular to it is `-b/a`.

As equation of one line is given in slope-intercept form as `y=-5/3x-8`, its slope is `-5/3`.

Hence slope of line perpendicular to it is `3/5`.

Now equation of line passing through `(x_1,y_1)` and having slope `m` is `(y-y_1)=m(x-x_1)`

Hence, equation of a line passing through `(-9,-3)` and having a slope `3/5` is

`(y-(-3))=3/5(x-(-9))`

or `5(y+3)=3(x+9)`

or `5y+15=3x+27`

or `3x-5y+12=0`
graph{(3x-5y+12)(y+5x/3+8)((x+9)^2+(y+3)^2-0.05)=0 [-20, 20, -10, 10]}