Line passes throught (2,2) and is perpendicular to y=3x+7

For this question you would need to use point-slope form which is: (y-y₁)=m(x-x₁) where m=slope and (x₁, y₁) is your point on the line. So you already have a point, (2,2) and you need slope which is the negative inverse of the slope of a line perpendicular to it. In this case, the slope would be -1/3. So if you plug all of this in the point-slope equation you get: y-2=(-1/3)(x-2) > y=(-1/3x)-(2/3)+2 >y=(-1/3x)+(4/3)