persamaan garis yg berbergradien -1/3 dan melalui titik (1,3) adalah...

y-y1=m(x-x1)
y-3=-1/3(x-1)
y-3=-1/3x+1/3
y=-1/3x+1/3+3
y=-1/3+10/3

y-y1 = m(x-x1)
y-3 = -1/3(x-1)
y-3 = -1/3x+1/3
-------------------- × 3
3y - 9 = -x + 1
3y + x - 9 - 1 = 0
x + 3y - 10 = 0