What is an equation of the line that is perpendicular to y+1=-3(x-5)

 3y - x - 42 = 0

Step-by-step explanation:

The equation                         y  +  1 = -3(x - 5)

Now expand the expression by opening the brackets

                                              y + 1 = -3x + 15

Recall,

Equation of line , y = mx + c , where m = slope

rearranging the equation above in order to determine the  ( m)

                                             y = -3x -1 + 15

                                             y = -3x + 14

Recall again  condition for perpendicularity

                                             m₁m₂ = -1

From the equation given, m₁ = -3, therefore , to get m₂

                                           -3 x m₂ = -1

                                            -3m₂    = -1

Therefore                               m₂    = -1/-3

                                               m₂   = 1/3

Now , to find the equation of the second line parallel to the first one,

Just replaced the -3 in the equation above by  m₂ = 1/3 in

               y = -3x + 14

               y = x/3 + 14

Now , multiply both the RHS and LHS of the equation by 3, to make it a linear.

              3y = x + 3 x 14

               3y = x + 42

              3y - x - 42 =0