Question

Plz help, 15 points!

What is the equation of a line that is parallel to 2x+3y=3 and passes through the point (3, −4)?


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_____

Answer 1

`2x+3y=-6`

Explanation:

Let the line is
`y=mx+b` where
`m` is slope and
`b` is
`y-intercept`.

The line is parallel to
`2x+3y=3`, slope of both the lines will be same.

Find the slope of
`2x+3y=3`

`3y=3-2x\\y=-(2)/(3)x+1`

Slope of line
`=-(2)/(3)`

`m=-(2)/(3)`

So the line will be
`y=-(2)/(3)x+b`

It passes through
`(3,-4)`.

`-4=-(2)/(3)* 3+b\\-4=-2+b\\b=-2\\`

Hence the line is
`y=-(2)/(3)x-2`

`y=-(2)/(3)x-2\\3y=-2x-6\ \ \ \ \ \ \ (Multiply both sides by 3)\\2x+3y=-6`