How to prove that there is a unique line between 2 distinct points using linear algebra?

Question

I want to prove using the parametric equation of line that any 2 distinct points $\ P,Q \in \mathbb{R}^2 $ is connected by a unique line. We know that a such a line can be formed by fixing one of the points as the end point of the starting vector and the other vector as a direction vector. We can represent the line as $ L = \{ P + \lambda (Q-P) \}$ or $ L = \{ Q + \mu (P-Q) \}$ where $\lambda,\mu \in \mathbb{R}$. How to prove that the two forms are equivalent to each other?

Answer 1

Let's write $L = \{ P + \lambda (Q-P) \}$ and $M = \{ Q + \mu (P-Q) \}$, so that what you want to prove is that $L = M$.

In general the trick for showing that two sets $L$ and $M$ are equal is separately showing the inclusions $L \subset M$ and $M \subset L$, or in other words: write one proof that every element of $L$ is also an element of $M$ and another proof that every element of $M$ is also an element of $L$.

Of course both proofs will be quite similar, but still having two of them is often easier than trying to do it in one fell swoop.

Answer 2

Notice that each dimension is independent, so we can perform the analysis for a single component $p \in \mathbb{R}$ and $q \in \mathbb{R}$.

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If $p=q$, the equivalence is trivial.

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If $p\neq q$, we can find a bijective function that relates $\mu$ and $\lambda$, so that any point that belongs to $L$ can be expressed by both forms. Which proves the equivalence:

$$p + \lambda\,[q-p] = q + \mu\,[p-q] \Leftrightarrow [p-q] = [\mu + \lambda]\,[p-q] \Leftrightarrow \boxed{\mu = 1 - \lambda}$$

To illustrate:

Let $x$ belong to the line. Then, there must be a value of $\lambda$ such that, $x=p + \lambda\,[q-p]$. Which means that we can obtain $\mu$ (as shown above) such that $x = q + \mu\,[p-q]$. So belonging to one form implies belonging to the other.