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Write the equation of the line that is parallel to x – 3y = 9 and passes
through the point (3, -1).


Sagot :

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]