Proof about field extension : A geometric way

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Let $M \subset \mathbb C $ be a sub-field which is not contained in $\mathbb R$ and which is closed under complex-conjugation.

Let $L(M)$ be the set of all lines which crosses two points of $M$ and let $C(M)$ be the set of all circles for which holds that their center is a point in $M$ and their radius is a absolute value of a point in $M$

Show the following statements:

(a) If z is the intersection of two lines from $L(M)$ then $z \in M$

Before posting task (b) and (c) I want to understand how to solve (a).

Unfortunately I got no idea how to solve it. I can't see any connection between that intersection point z and the set L(M).

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Because this is a homework I give you only few hints:

  1. If $x+iy\in M$, then $x-iy\in M$ and therefore $x\in M$, $iy\in M$.

  2. Take four points $z_k=x_k+iy_k\in M$ and a point $p=x+iy$ that lies on the lines determined by $z_1,z_2$, and $z_3,z_4$, respectively. Now write the equations of both lines and solve them for $x$. In the end you get a linear equation in $x$ with coefficients in $M$, so $x\in M$. Analogously you get $iy\in M$.