Question

What is the equations of a line that is parallel to y = -3x + 6 and contains the point (-3,-5)?

Answer 1

`y=-3x-14`

Explanation:

Any line parallel to `y=-3x+6`
will be in the form `y=-3x+c` for some constant `c`

If `(x,y)=(-3,-5)` is a solution to such an equation, then replacing `y` with `(-5)` and `x` with `(-3)` gives
`-5=(-3) *(-3)+c`

`rarr -5=9+c`

`rarr c=-14`

So the desired parallel line would have the equation
`y=-3x-14`

Answer 2

The equation of the line parallel to `y=-3x+6` that passes through the point `(-3,-5)` is:

`y=-3x-14`

Explanation:

The line `y=-3x+6` is already in 'slope-intercept' form, so we can just read off the slope (gradient) as being `-3`.

We want a line with a gradient of `-3` (because that's what 'parallel' means - having the same gradient) that passes through the point `(-3,-5)`.

Take the form `y=mx+c` and substitute in the 3 things we already know: the slope, one value of `x` and one value of `y`:

`-5=(-3)(-3)+c`

`c` is the y-intercept of the line: the `y` coordinate of the point at which it cuts the `y` axis (which is the line `x=0`).

`-5=9+c`

Subtract `9` from both sides:

`-14=c`

That means the equation of the line is `y=-3x-14`