If $x,y\in \mathbb R^n$, then $tx+(1-t)y$ for any $t\in R$ lies on a 'line'. What is line here?

Question

I was reading Generalized MVT. I got stuck in the following concept.

If $x$ and $y$ belong to $ \mathbb R^n$ then it is said that $tx+(1-t)y$ for any $t\in R$ lies on a 'line'. What is line here? How $tx+(1-t)y$ for any $t\in R$ belongs to that line?

Can anyone please help me out?

Answer 1

Let $x,y\in\mathbb R^n$ be two distinct points (or vectors) in the Euclidean space.

A line passing through the points $x$ and $y$ is defined to be the set of points $$ \ell(x,y)=\left\{tx+(1-t)y\mid t\in\mathbb R\right\}$$

If the restriction $0\leqslant t\leqslant 1$ be imposed on $t$, then the point $tx+(1-t)y$ on this line is constrained to lie within the segment joining the points $x$ and $y$. Thus the set

$$\ell'(x,y)=\{tx+(1-t)y\mid 0\leqslant t\leqslant 1\}$$ is defined to be the line segment joining the points $x$ and $y$.

Indeed, the line segment is nothing but the set of convex combinations of $x$ and $y$.

Answer 2

Write down a parametric equation for the straight line passing through $x$ and $y$, then you will see.

UPDATE

The line in $\mathbb R^n$ is the set defined in your text, which is an analogy of simple cases $n =2,3$.

In $\mathbb R^2, \mathbb R^3$, a line is a subset congruent to a 1-dimensional subspace of the whole space. Generally we could also give the same name to such kind of subsets in $\mathbb R^n$.

In $\mathbb R^n$, given two fixed points $x,y$. If $L$ is a subset congruent to $\mathbb R^1$, then for each $z \in L$, $z=y +t(x-y)$ for some $t$, or $z = tx + (1-t)y$. So we define $$ L = \{z \in \mathbb R^n \colon z = tx +(1-t)y \;[t \in \mathbb R]\} $$ to be the line passing through $x,y$.

Since in $\mathbb R^3, \mathbb R^2$, whenever $t \in [0, 1]$, $z = tx+(1-t)y$ lie in the segment determined by $x,y$, and $z$ lies "between" $x,y$, we analogously call $L_{t \in [0,1]}$ the segment $[x,y]$.