Question

How do you convert `(y = 2 + x)` to parametric equation?

Answer 1

`l->{(x = 0+lambda),(y = 2+lambda):}` for `lambda in RR`

Explanation:

The general line equation in `RR^2` is

`l->ax+by+c=0`

this equation can be arranged as

`l-> << vec n, p-p_0>> = 0` where

`vec n = {a,b}`, `p = {x,y}` a line generic point, and `p_0={x_0,y_0}` a line given point.

The determination of `p_0` is straightforward,
comparing

`ax+by+c=0` with
`a(x-x_0)+b(y-y_0) = 0` concluding

`-a x_0-b y_0 = c`

then supposing that `b ne 0` and fixing `x_0=0` we get `y_0=-c/b` so

`p_0={0,-c/b}`

Now, the parametric representation.

We have a line point which is `p_0` and an orthogonal vector `vec n` to the line direction. We need a vector `vec v` in the line direction, so choosing `vec v` such that `norm(vec v) > 0` and `<< vec n, vec v>> = 0`. Choosing `vec v = {b,-a}` then `<< vec n, vec v>> = b a-a b = 0`.

Finally, the parametric line equation is

`l->p = p_0 + lambda vec v` with `lambda in RR`

In the present case

`l->-x+y-2=0`

`p_0 = {0,-c/b} = {0,-((-2)/1)}`
`vec n = {-1,1}->vec v = {1,1}`

then

`l->{(x = 0+lambda),(y = 2+lambda):}`

Answer 2

see below

Explanation:

parametrising a straight line is a bit of a waste of time unless you have an idea as to why you would want to do that.

or, put another way, the idea behind parameterisation is that you reduce the independent variables. in Cartesian x-y, it really kinda misses the point.