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