Vtyjkyuk

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 $tin R$ lies on a 'line'. What is line here? How $tx+(1-t)y$ for any $tin R$ belongs to that line?

Can anyone please help me out?

Let $x,yinmathbb R^n$ 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 $$ ell(x,y)=lefttx+(1-t)ymid tinmathbb Rright$$

If the restriction $0leqslant tleqslant 1$ 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

$$ell'(x,y)=tx+(1-t)ymid 0leqslant tleqslant 1$$ 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 $y$.

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 $[x,y]$.