Show, three different points $z_1,z_2,z_3\in\mathbb{C}$ are on a straight line if and only if there is a real number $r$ with $z_3-z_1=r(z_2-z_1).$

complex numbersgeometry

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:

  • if you take $t=0$, you deduce that $z_1\in R$;
  • if you take $t=1$, you deduce that $z_2\in R$;
  • if you take $t=r$, you deduce that $z_3\in R$.

Can you take it from here?

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