Question 107079
the line y=1/3x+5 has a gradient of 1/3 (the coefficient of the x value when y is the subject of the formula)
the line perpendicular is going to have -3/1 as the gradient (if two line are perpendicular the gradients are the negative recipricol of each other so I flipped 1/3 up-side-down and stuck a negative infront. 

Now you have the gradient of the new line and a point, you use the point-gradient formula

y-y1=m(x-x1)

where x1 and y1 are the coordinates you are given (x1=1, y1=4) and m=-3/1=-3

subbing in:
y-4=-3(x-1)
y-4=3x+3
y-3x-7=0