Question

What is the equation of the line perpendicular to 3x+4y=12 and containing the point (6,-2)

Answer 1

3x+4y=12

4y = - 3x + 12

 y = -3/4 x + 3; This line has slope equal - 3/4

Perpendicular lines, slope is opposite and reciprocal so slope of new line = 4/3

Passing point (6 , -2)

y + 2 = 4/3(x - 6)

y + 2 = 4/3 x - 8

y = 4/3 x - 10

Answer

Equation: y = 4/3 x - 10