Question

What is the equation of the line that is perpendicular to the line y=3/5x+10 and passes through the point (15, -5)?

Answer 1

Perpendicular lines have slopes that are negative reciprocals, so the slope of the line we want to find is -5/3.

Substituting into point-slope form,

$y+5=-\frac{5}{3}(x-15)\\\\y+5=-\frac{5}{3}x+25\\\\\boxed{y=-\frac{5}{3}x+20}$