Question

How do you write the equation given (-6,7); parallel to 3x + 7y = 3?

Answer 1

`s: y = -3/7 x + 31/7`

Explanation:

We take the line above in the form `r:y = ax + b`
We know that any parallel is `s:y = ax + c`
We choose `(-6, 7) in s`

`7y = 3 - 3x => r: y=-3/7x+3/7`

`s:7 = -3/7 (-6) + c`

`49 = 18 + 7c => c = 31/7`

Answer 2

`y=-3/7x+31/7`

         or

`7y=-3x+31`

Explanation:

Change `3x+7y=3` to standard form of `y=mx+c`
`7y=3-3x`
`y=-3/7x+3/7`

gradient, `m`, can be determined as `-3/7`

Two parallel lines would have the same gradient, in this case gradient of `-3/7`

You can choose to use gradient formula
Gradient, `m=(y_1-y_2)/(x_1-x_2)`
or general formula for straight line
`(y=mx+c)`

I would first be attempting it using gradient formula

replace `m` with `-3/7`; replaace `x_1,x_2,y_1,y_2` with `x` and the x-coordinate, `y` and the y-coordinate respectively

in this case, you have only 1 point given, if there is more, the x and y coordinate must be from the same point.

`-3/7=(y-7)/(x-(-6))`
`-3/7=(y-7)/(x+6)`

rearrange
`-3/7(x+6)=(y-7)/(x+6)`
`-3/7x-18/7=y-7`
`y=-3/7x-18/7+7`
`y=-3/7x+31/7`

in case you don't like fractions, multiply whole equation by 7
`7y=-3x+31`

Using general formula

replace `m` with `-3/7`; `x` with the x-coordinate, `y` with the y-coordinate

in this case, you have only 1 point given, if there is more, the x and y coordinate must be from the same point.

`7=(-3/7)(-6)+c`

solve for c
`7=18/7+c`
`c=31/7`

substitute `m` and `c` into `y=mx+c`
`y=-3/7x+31/7`
`7y=-3x+31`

Both equation for straight line you get are the same, depending on which you prefer.