Answer:
x – 3y = 9 ⇔ y = [tex]\frac{1}{3} x + 3[/tex]
the line parallel is y' = [tex]\frac{1}{3}x + b (b\neq 3)[/tex]
y' = [tex]\frac{1}{3}x + b (b\neq 3)[/tex] passes through (3, -1) --> -1 = [tex]\frac{1}{3}[/tex] × 3 + b --> b = -2
⇒ y' = [tex]\frac{1}{3} x - 2[/tex]
