find the equation of the line perpendicular to the graph of x+3y=-3 that contains the point (-2,-2)

Answer 1

x+3y=-3

3y = -x - 3

y = -(1/3)x - 1

perp lines have slopes that are opposite in sign and the inverse of each other.

So, the new slope will be 3

y = 3x + b

(-2,-2)

-2 = 3(-2) + b

-2 = -6 + b

b = 4

y = 3x + 4