If and only if then means that I have to show an equivalence. I could prove one implication, but not the other. The solutions says:
A relation
$z_3-z_1=r(z_2-z_1)$
with $r\in\mathbb{R}$ is equivalent to say the vectors $z_3-z_1$
and $z_2-z_1$ are $\mathbb{R}-$linear dependent
Now I have understood that $z_3-z_1$ and $z_2-z_1$ are
vectors and that they are also linearly dependent. But how does this answer the question? Where is the straight line and what can I say about the points $z_1,z_2,z_3$?
Here are my results so far:
Suppose $z_1,z_2,z_3$ are on a straight line.
We have to show that the points are
$z_3-z_1$ and $z_2-z_1$ are on the same straight line as the points just mentioned.
The following applies $z_3=az_1$
and thus $z_3-z_1=az_1-z_1=(a-1)z_1$
So the point $z_3-z_1$ also lies on the straight line.
You can argue similarly with the other point. So both points are on the straight line and because of this a $r\in\mathbb{R}$ also exists so that
$r(z_2-z_1)=z_3-z_1$
applies.
I could not prove the other implication
My thoughts so far on this are:
If
$r(z_2-z_1)=z_3-z_1$
then the points must be
$z_2-z_1$ and $z_3-z_1$
on a straight line. If you add location vectors of points of a straight line you get a location vector for a new point on the same straight line.
So I added the two location vectors and got :
$z_3-z_1+z_1-z_2=z_3-z_2$
So the point $z_3-z_2$ is on the straight line again.
I then continued to add the location vectors of the 3 points I have so that I can show that at least one of the points $(z_1,z_2,z_3)$ can be represented as a sum.
But I didn't make it, I looked into the solutions and expected to find the identity
$r(z_2-z_1)=z_3-z_1$
I took advantage of him.
My approach must have been a wrong one, I hope someone can explain the solution to me so that I can at least comprehend this implication.
Best Answer
If there is a real number $r$ such that $z_3-z_1=r(z_2-z_1)$, then consider the straight line$$R=\{z_1+t(z_2-z_1)\,|\,t\in\mathbb{R}\}.$$Then $z_1,z_2,z_3\in R$. In fact:
Can you take it from here?