Ah, I see. He's starting with a parametric equation for a line:
\left(\begin{array}{c}x \\ y\end{array}\right) =<br />
\left(\begin{array}{cc} 2-\alpha\\\alpha \end{array}\right) =\left(\begin{array}{cc} 2\\0\end{array}\right) +\alpha\left(\begin{array}{cc} -1\\1\end{array}\right)
and he wants to know how to turn it into an implicit form. (i.e. the solutions to an equation)
There's a simple approach that works: just take the equations
x = 2 - \alpha
y = \alpha
and eliminate the variable \alpha.
There's a highbrow approach too, that (IMHO) is a good example to learn.
To simplify things, let me write the original equation as:
\vec{z} = \vec{b} + \alpha \vec{v}
In order to get rid of \alpha, you need to multiply by something that kills off \vec{v}: you want a good matrix A such that A \vec{v} = 0. Well, that's just a left-nullspace matrix computation! (i.e. the columns of A^T are a basis for the nullspace of \vec{v}^T)
In this case, we get something like
A = \left(\begin{array}{cc}1 & 1\end{array}\right)
And so the desired equation is
A \vec{z} = A \vec{b}
which you can check reduces to x + y = 2.
