The line perpendicular to 3x+y = 1 is
x - 3y = c, (1) where c is an arbitrary constant.
To specify the perpendicular line through the point (-2,1), put in equation (1)
x= -2, y= 1. You will get then
c = -2 -3*1 = -2 - 3 = -5.
Thus the equation you are seeking for is x - 3y = -5. ANSWER