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What is the slope of a line perpendicular to the line whose equation is

3x−12y=−108. Fully simplify your answer

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Answer:

Explanation:

The given equation is not instandard form. First standard form is neccesary.

  • Add 108 to both sides.
  • Subtract 12y to both sides

The result is 3x + 108 = 12y which is equal to 12y = 3x + 108

The slope is represented in this form as m (Y = mx + b)

(with b as the initial value)

Therefore, 3 is the slope.

To find its perpendicular slope.

To find it's perpendicular, know that it is the negative reciprocal.

  • First negate 3 to become -3 and find it's reciprocal
  • To find the reciprocal, divide 1 by -3

FInally, the slope perpendicular to the line whose equation is 3x - 12y = -108 is
(1)/(-3\\)

User Jmrah
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