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

Answer: 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