$y = \theta x_1 + (1 − \theta)x_2$

Question

I am following Prof. Boyd’s lecture on Convex optimization. In the second lecture it is mentioned that the eq. $y = \theta x_1 + (1 − \theta)x_2$ represents the points lying on a line joining $x_1$ and $x_2$. I am unable to find a proof for this.

I tried to solve this taking a generic point $x_3=a \cdot x1+b\cdot x_2$ and solving for $dist(x_1 \to x_3) + dist(x_3 \to x_2) = dist(x_1 \to x_2)$, to get $a+b=1$. But the equation gets complex.

Following are course links: http://stanford.edu/class/ee364a/lectures.html

If someone can give a proof for this or give a reference which gives the proof.

score 0 · Answer 1 · answered Jul 12 '17 at 06:32

0

You will agree that $$y-x_2=\theta(x_1-x_2)$$ is a line through the origin, won't you ?

Then $y=(y-x_2)+x_2$ is a line through $x_2$, parallel to $\vec{x_2x_1}$.

answered Jul 12 '17 at 06:32

score 0 · Accepted Answer · answered Jul 12 '17 at 06:33

Well, you have $\theta x_1+(1-\theta)x_2=x_2+\theta(x_1-x_2)$, so all these poits belong to the straight line passing through $x_2$ with the direction given by the vector $x_1-x_2$. Furthermore, it is clear that, as $\theta$ goes from $0$ to $1$, $\theta x_1+(1-\theta)x_2$ goes from $x_2$ to $x_1$.

$y = \theta x_1 + (1 − \theta)x_2$

2 Answers2