Find the equation for the line that passes through 5he point (5,1) and is perpendicular to the line x-3y+16=0

please show steps. Any help appreciated!

-3y = -x - 16
y = x/3 + 16/3

slope = 1/3
slope of perpendicular line = -3

y - 1 = -3(x - 5)
y - 1 = -3x + 15
y = -3x + 16

The line x-3y+16=0 has a slope =1/3
So a line eprpendicular to it must have slope = (-3)
Knowing slope and one point (5,1), we can write the equation of the perpendicular line as (y-5)=(-3)(x-1)
or y=(-3)x+8

x-3y+16=0
3y = x+16
y = (x+16)/3

the slope of this line = 1/3

the slope the prependicular = -1/the slope = -3

the equation is

(y-1) = -3 (x-5)
y-1 = -3x + 15
y = -3x + 16

-3y=-x-16 , m=-1/-3=1/3
m_|_=-3
(y-1)/(x-5)=-3
y-1=-3x+15
y=-3x+16
God bless you.