Why is that linear combinations of coefficients with sum 1 give a line?

Question

In case of $$p = t\cdot p_1 + (1 - t) \cdot p_2, \text{ for } p_1, p_2 \in \Bbb R^2, t \in [0, 1]$$ the end of vector $p$ will always land on the line between the endpoints of $p_1$ and $p_2$. Is there an $intuitive$ way to describe this phenomenon?

Answer 1

I'll rename the points $p_0$ and $p_1$ to make the description look a bit nicer.

The line through the points $p_0$ and $p_1$ is parametrized by some linear function $$p_t=at+b,$$ for some real numbers $a$ and $b$, with $t\in\Bbb{R}$. Putting $p_0$ at $t=0$ and $p_1$ at $t=1$, the points between $p_0$ and $p_1$ are parametrized by $t\in[0,1]$. Plugging in $t=0$ and $t=1$ we get $$a\cdot0+b=p_0,$$ $$a\cdot1+b=p_1,$$ and hence $b=p_0$ and $a=p_1-p_0$. It follows that $$p_t=(p_1-p_0)t+p_0=t\cdot p_1+(1-t)\cdot p_0.$$ And as noted before, the points between $p_0$ and $p_1$ are parametrized by the values of $t$ between $0$ and $1$.

Answer 2

$p_i$ are vectors. You add vectors by putting the tail of one on the origin and lining up its head with the other's tail. The result is where the head of the second lands. $t$ is rescaling the length of each vector, saying you want some fraction of $p_1$ and the remaining portion of $p_2$. If you take say, $1/4$ of $p_1$ and $3/4$ of $p_2$, then the direction will be $1/4$ of the way from $p_1$ to $p_2$ and the length will be $|p_1|/4+3|p_2|/4 $ where $|\cdot|$ represents length. In a sense you are weighting the direction and the length. If you consider all the possible lengths and all the possible angles you can get from putting weights on each you might think it would make a rectangle (that any length from the length of the shorter vector to the length of the longer vector could be at any position between the endpoints of the two vectors), but if you are weighting toward 1 of them more than the other it will affect both position and length.