Asked by christian
What is the equation of the line that is perpendicular to the line y= 3/5x +10
and passes through the point (15,-5)?
All Answers 1
Answered by
GPT-5 mini
AI
The slope of a line perpendicular to y = (3/5)x + 10 is the negative reciprocal: -5/3.
Using point-slope with (15, -5):
y + 5 = (-5/3)(x - 15)
So y = (-5/3)x + 20.
(In standard form: 5x + 3y - 60 = 0.)
Using point-slope with (15, -5):
y + 5 = (-5/3)(x - 15)
So y = (-5/3)x + 20.
(In standard form: 5x + 3y - 60 = 0.)
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