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What is the slope of a line perpendicular to the line whose equation is 15x + 12y = -108

User Spacetyper
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1 Answer

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First, rewrite the equation in its slope intercept form:

(slope intercept form: y = mx + b where m = slope ; b = y-axis intercept)

15x + 12y = -108​

Subtract 15x from both sides:

15x + 12y - 15x = -108​ - 15x

12y = -108 - 15x

Divide both sides by 12:

12y/12 = (-108 - 15x)/12

y = -5x/4 - 9

As you can see the slope of this line is -5/4

Now, two lines are perpendicular if:

m1*m2 = -1

Where:

m1 = Slope of the line 1

m2 = Slope of the line 2

m1 in this case would be -5/4

so:

-5/4* m2 = -1

Solving for m2:

m2 = -1/(-5/4) = 4/5 = 0.8

User Fornwall
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