Question

What is the equation of a line that passes through the point (6,5) and is perpendicular to the line 3x+4y=40

Answer 1

3x + 4y = 40

4y = -3x + 40

y = (-3/4)x + 10

A line perpendicular to this line will have slope 4/3.

y - 5 = (4/3)(x - 6)

y - 5 = (4/3)x - 8

y = (4/3)x - 3

3y = 4x - 9

-4x + 3y = -9

4x - 3y = 9