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.)