Question

How do you write an equation in standard form for a line passing through (–4, 2) and is parallel to y = –x + 6?

Answer 1

All lines parallel to `y=ax+b` have to have the same coefficient at `x` in the form `y=px+q`. That is, `p=a`.

In our case all lines parallel to `y=-x+6` have to have a form `y=-x+q`, where `q` can be any real number.

The condition that our line should pass through `(-4,2)` results in the equation, from which we can determine an unknown variable `q`:
`2=-(-4)+q`
`q=-2`

Therefore, the equation of a line we are looking for is
`y=-x-2`

It passes through `(-4,2)` since, if we substitute `-4` for `x`, the value of `y` will be `2`.
It also parallel to `y=-x+6` because the coefficients at `x` are the same in both cases.

Graphs of these two equations are:
`y=-x+6`
graph{-x+6 [-10, 10, -5, 5]}
`y=-x-2` - notice the line is passing point `(-4,2)` and is parallel to `y=-x+6` above
graph{-x-2 [-10, 10, -5, 5]}