# Probably very basic Euclidean geometry; Why is the following expression valid for a point along a straight line?

I am looking at constructible points in abstract algebra, particularly in $$\mathbb{C}$$. Alongside a proof of a theorem, I came across this expression which I cannot work out how it's been derived. It just comes out as follows, but some notations;

$$L(z_1,z_2)$$ represents the straight line that connects points $$z_1,z_2 \in \mathbb{C}$$. $$\mathbb{Q}^{Py}$$ represents the Pyhtagorean Closure of $$\mathbb{Q}$$.

The theorem is stated as follows

A point $$z \in \mathbb{C}$$ is constructible if and only if $$z \in \mathbb{Q}^{Py}$$.

Omitting the bits I understood(The proof uses induction), the specific part I don't understand is,

Say $$z$$ lies on a "line meeting another line", namely $$\{z\} \in L(z_1,z_2) \cap L(z_3,z_4)$$ where the lines are distinct. Then $$\exists a,b \in \mathbb{R}$$ such that

$$z=az_1+(1-a)z_2$$

$$z=bz_3+(1-b)z_4$$

Well, in all honesty, it looks familiar; points on a line that goes through $$2$$ distinct points. I think the above expressions come from rather elementary facts and ideas but as much as I am embarrassed to say this, I can't see how it gets derived. Considering the Complex plane as $$\mathbb{R}^2$$, taking $$x,y$$ coordinates, I thought I could find $$a$$ or $$b$$ wrp to $$z_j=x_j+iy_j$$ but I cannot.

Would someone show me how that part is derived?

You have the right idea, you should be considering it as $\mathbb{R}^2$. The line $L(z_1,z_2)$ can be defined as the image of the straight-line path between its endpoints:

$$f : [0,1] \rightarrow \mathbb{C} \\ f(t) = z_1 + t(z_2 - z_1) \\ = tz_2 + (1-t)z_1$$

So any point on the line can be represented as $f(t)$ for some $t$. The step you're looking at is just choosing $a$ and $b$ to be the appropriate values for the point being considered, in terms of the two different lines it is on.

• I replaced the idea with vectors and I think I got it; I arrived at the same form as yours up there, thanks for the help! – John Trail Feb 5 '16 at 18:39

Rewrite $z = a z_1 + (1 − a) z_2$ as:

$$a ( z - z_1) + (1 - a) (z - z_2) = 0$$

Since $a \in \mathbb R$ this means that $z - z_1$ and $z - z_2$ have a real ratio, which happens iff $z$ is on the $z_1 z_2$ line.