Expressing a line as a linear combination of two points on the line. I'm currently reading Pugh's Analysis. He makes the statement that the line between two points x and y is the set of linear combinations $sx + ty$ where $s + t = 1$. I'm satisfied that this is true, as the line between two points $x$ and $y$ has the equation $y-x_2=\frac{x_2-y_2}{x_1-y_1}(x-x_1)$ and the points of form  $(sx_1+ty_1, sx_2+ty_2)$ are solutions of the equation. I still don't have any reasonable geometric intuition for why this is true though. Help?
 A: It’s easiest to see what’s going on when $0\le s,t\le 1$. In that case $sx+ty$ is a weighted mean of $x$ and $y$. When $s=t=\frac12$, for instance, it’s the ordinary arithmetic mean, and the point $\frac12x+\frac12y$ is the midpoint of $\overline{xy}$. When $s=\frac13$ and $t=\frac23$, $y$ is given twice the weight of $x$, so the point is ‘twice as close’ to $y$ as it is to $x$: in more understandable terms, it’s half as far from $y$ as it is from $x$, so that it’s $\frac23$ of the way from $x$ to $y$. In general, $sx+ty$ is $t$ fraction of the way from $x$ to $y$ when $0\le s,t\le 1$.
Once $s$ and $t$ get outside the range $[0,1]$, it’s harder to get an intuitive picture, but you can still regard $sx+yt$ is a kind of weighted mean in which one of the weights is negative. Thus, $-x+2y=y+(y-x)$ is the point that is $y-x$ units past $y$ in the direction away from $x$, and in general $sx+ty$ is the point that is $y-x$ units past $(t-1)y$ in the direction away from $x$ when $t>1$. (And of course everything just turns around when $s>1$.)
A: I think the easiest geometric explanation comes from this set of pictures:

So, the general equation for all linear combination of the points, is as follows:
$\vec{y} = \vec{x}_2 + \alpha(\vec{x}_1-\vec{x}_2) = \alpha\vec{x}_1 + (1-\alpha)\vec{x}_2$, $\alpha \in R^n$
if we name, $s=\alpha$ and $t=1-\alpha$.  Then,
$\vec{y} = s\vec{x}_1 + t\vec{x}_2$
A: I know this is really late but I am going to try to answer this anyways. I had the same problem visualising this. 
The easiest way to understand this (for me personally) is to use homogenous co-ordinates. We use the implicit definition of a line $L$ as the set of points which satisfy $ax + by + c = 0$. So we can represent a line a $3 \times 1$ vector of the co-efficients $(a,b,c)'$ where $'$  denotes transposition.
Any point $p$ which satisfies $p'\cdot L = 0$ is on the line. since $x_1'\cdot L= 0$ and $x_2' \cdot L=0$ any linear combination of the 2 will also be zero and hence be on the line $L$.
Hope this helps.
