A line is parametrized by: $ \mathcal{L}_{\bar{u}}(\bar{p}) = \{ \bar{p} + r\bar{u} \}.$
In an informal conversation I had to show that $ \mathcal{L}_{\bar{v} - \bar{u}}(\bar{u})$ implies $ r + s = 1$ for all $\bar{v} , \bar{u}.$
It is simple, as $ \mathcal{L}_{\bar{v} - \bar{u}}(\bar{u}) = \{ \bar{u} + r(\bar{v} - \bar{u}), r \in \mathbb{R} \} = \{ \bar{u} + r\bar{v} - r\bar{u} \} = \{ (1 -r)\bar{u} + r\bar{v} \} . $ Let $s = 1 - r.$ Then $ s \in \mathbb{R}$, and $ r + s = 1.$
My question is: What is the more complete/correct formulation of this question? I guess it is an if and only if proof statement, but what would be the statement, and how is this observation important or useful?
If L is the line passing through $\bar {u}$ in the direction of $\bar {v} - \bar {u}$ then a point $\bar {z}$ lies on this line if and only if $\bar {z}=r\bar {u}+s\bar {v}$ for some r and s with $r+s=1$.