Intuition behind the formula of convex combination of two distinct points

Question

A convex combination of two vertices $p = ( a, b )$ and $q = ( c, d )$ is any point $r = ( e, f )$ such that for some $x$ in range $0 \le x \le 1$, $e = xa + ( 1 - x )c$ and $f = xb + ( 1 - x )d$. Intuitively, $r$ is any point that is on the line passing through $p$ and $q$ and is on or between $p$ and $q$ on the line.

Is there any intuition about why this combination is always going to lie on the line from $p$ to $q$?

The question is different from If $0 \le a \le 1$, then show that $xa+(1−a)y$ will always lie between $x$ and $y$. as if e lies between a and c and f lies between b and d it does not gaurantee that it will be on the line. For example let $p$ be (1,2) and q be (3,4) and r be (2.5,3). It follows the above constraint but yet it does not lie on the line.

Proof

That said, can now somebody give intuitive proof of this now ?

Answer 1

In your example $p = (1,2)$ and $q = (3,4)$, $r = (2.5,3)$ is not a convex combination. By solving for $x$, in the first coordinate, we get $$3x - 1(1-x) = 2.5 \Rightarrow x = \frac{7}{8}$$ but in the second coordinate $$ 4x - 2(1-x) = 3 \Rightarrow x = \frac{5}{6} $$ It follows that $r \notin \text{conv}(p \cup q) = \lbrace s : s = \lambda p + (1 - \lambda)q, 0 \leq \lambda \leq 1 \rbrace$

I think this should clear up the confusion.

Answer 2

Think of $X = x p + (1-x) q = q + x (p - q)$ as the parametric equation of a point $X$. It is obtained by starting at $q$ and moving parallel to the vector $p-q$. Thus this is the parametric equation of a straight line. At $x = 0$ you get $X=q$, at $x=1$ you get $X=p$, so it's the straight line through points $p$ and $q$. As $x$ goes from $0$ to $1$, the point $X$ moves in the direction of $p-q$ from $q$ to $p$, thus traversing the line segment from $q$ to $p$.