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Write an equation of a line that is parallel to the line whose equation is 3y=x+6 and that passes through the point (-3,4).

Sagot :

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The standard form of this equation is

[tex]y=\frac{1}{3}x+2[/tex]

The slope of two parallel lines i the same, so the slope of the line we're looking for is also [tex]\frac{1}{3}[/tex]

Since the line passes through [tex](-3,4)[/tex], the equation is:

[tex](y-4)=\frac{1}{3}(x+3)[/tex]

[tex]\boxed{y=\frac{1}{3}x+5}[/tex]





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