Justifying the analytic definition of a line segment

Question

The definition of a line segment $L$ for $x$ and $y$ in a vector space is $$L = \{\lambda x+(1-\lambda)y : \lambda \in [0,1]\}.$$

I had trouble seeing this, so I considered the basic case of $\mathbb{R}$ equipped with the usual metric.

Expanding I got $$\lambda x + y - \lambda y = \lambda(x-y) + y$$ Here my intuition can work. The difference $x-y$ has a distance and we are going to shorten it by $\lambda$, then add $y$ so that we get a point that is between $x$ and $y$ in the number line.

Now while that is nice and all for me, how could I reword that with more proper terminology?

I am adding the soft-question tag, because this can get a bit subjective.

Answer 1

Let $x,y \in \mathbb{R}^{n}$. Starting at $x$, the line $L$ parallel to $y-x$ through $y$ takes the form $x + t(y-x)$ where $t \in \mathbb{R}$. So the line segment joining $x$ and $y$ is simply the subset $\{ x + t(y-x) \mid t \in [0,1] \}$ of $L$. For all $t \in [0,1]$ we have $x + t(y-x) = (1-t)x + ty$. So we obtain the

definition. If $x,y \in \mathbb{R}^{n}$, then the line segment joining $x$ and $y$ is defined as the set $\{ (1-t)x + ty \mid t \in [0,1] \}$.

Answer 2

My intuition is that there is a starting point ($y$ in this case) and a direction ($x-y$ in this case).

The line starts at the starting point and moves in the direction. The amount of motion is specified by a parameter ($\lambda$ in this case).

Answer 3

Another way of looking at it is to view the point with position vector $(1-\lambda)x+\lambda y$ as the point which divides the line segment $XY$ in the ratio $\lambda:(1-\lambda)$