every line contains exactly $2$ points $\Bbb R^2=\{(x,y)\mid x,y\in\Bbb R\}$. Does there exist a set $S\subset \Bbb R^2$ such that every line $T_{a, b,c}=\{(x,y)\in\Bbb R^2\mid ax+by+c=0\}$, where $a$ and $b$ are not both zero, contains exactly $2$ points in $S$?
 A: Yes, see this article. And here are some more characters.
A: It’s easy to construct such a set. Let $\mathscr{L}=\{L_\xi:\xi<2^\omega\}$ be the set of straight lines in the plane. For each $A\subseteq\Bbb R^2$ let $\mathscr{L}(A)=\{L\in\mathscr{L}:|L\cap A|\ge 2\}$.
We’ll construct sets $A_\eta$ for $\eta<2^\omega$ so that $|A_\eta|\le\omega+|\eta|$, $A_\xi\subseteq A_\eta$, and $|L_\xi\cap A_\eta|=2$ whenever $\xi\le\eta<2^\omega$. Let $x_0$ and $y_0$ be distinct points of $L_0$, and let $A_0=\{x_0,y_0\}$. If $\eta<2^\omega$, and the sets $A_\xi$ for $\xi<\eta$ have been constructed, let $A_\eta'=\bigcup_{\xi<\eta}A_\xi$. If $L_\eta\in\mathscr{L}(A_\eta')$, let $A_\eta=A_\eta'$. If not, there are two cases to be distinguished.


*

*If $L_\eta\cap A_\eta'=\varnothing$, let $x_\eta$ and $y_\eta$ be distinct points of $L_\eta\setminus A_\eta'$; this is possible, since $|A_\eta'|<2^\omega=|L_\eta|$. Let $A_\eta=A_\eta'\cup\{x_\eta,y_\eta\}$.  

*If $L_\eta\cap A_\eta'=\{x\}$, let $x_\eta$ be any point of $L_\eta\setminus A_\eta'$, and let $A_\eta=A_\eta'\cup\{x_\eta\}$.


In all cases $|A_\eta|\le\omega+|\eta|$, $A_\xi\subseteq A_\eta$ for each $\xi\le\eta$, and $|L_\xi\cap A_\eta|=2$ for each $\xi\le\eta$, so the construction goes through to $2^\omega$. Let $A=\bigcup_{\xi<2^\omega}A_\xi$; clearly $L_\xi\cap A=L_\xi\cap A_\xi$ for each $\xi<2^\omega$, so each line intersects $A$ in exactly two points.
