Find the equation of the line that is perpendicular to 3y-x=7 and passes through the point (2,-1

Hello
an equation for the line in point-slope form and general form is :
y = ax+b   a : slope  ; the Passing through (x' ; y')
y' = ax'+b 
y-y' =a(x-x')      and : x' =2   y' = - 1
calculate a :
let :  y = ax+b (D)
   3y-x=7(D') or  y = (1/3)x+7/3
.(D) perpendicular to(D') : slope (D) × slope (D') = -1 

 slope (D') = 1/3
slop(D)×(1/3) = -1
slope (D) = -3
equation for the line    :   y-y' =a(x-x') 
y+1 =(-3) (x-2)

40 & 50 in the right corner are the complementary angles as when placed in a sum (added together) they make 90°