Every line in $\mathbb{R}^2$ can be described... I came across such a statement:

Let $A = \mathbb{R}^2$, $a,b,c \in A$ be points (we treat $\mathbb{R}^2$ as an affine space). Then any line $L \in A$ can be described as 
  $$L = \lbrace s_1 a + s_2 b + s_3 c \in A \mid s_1 + s_2 + s_3 = 1 \land c_1 s_1 + c_2 s_2 + c_3 s_3 = 0 \rbrace $$
  for some $c_1, c_2, c_3$. 

I can't find out, why this statement is true. Can you help me please?
 A: Let me change the notation slightly for convenience: The points I will call $A_1, A_2, A_3$ instead of $a,b,c$ (capital letters are always points in what I write). Also I will define
$$
L_{\mathbf{A}, \mathbf{n}} = \{
\mathbf{s}\cdot \mathbf{A}|\sum s_i = 1, \mathbf{n}\cdot \mathbf{s}=0
\}
$$
where $\mathbf{s}, \mathbf{n}$ are vectors in $\mathbb{R}^3$ and $\mathbf{s}\cdot \mathbf{A}$ is purely formal and means $s_1 A_1+s_2A_2+s_3A_3$. You need extra conditions:


*

*First of all you need $\mathbf{n}\neq 0$ that is necessary! If that is not the case then for example for $A_1=(1,0)$, $A_2=(0,1)$, $A_3=(0,0)$ you get the whole $\mathbb{R}^2$, and $L_{\mathbf{A}, 0}$ is not a line.

*The points $A_1, A_2, A_3$ should not be colinear (and not equal), i.e. they should not lie on a line themselves. If they are colinear then $A_3=A_1+\alpha(A_2-A_1)$ and $L_{\mathbf{A}, \mathbf{n}}$ for any $\mathbf{n}$ is (at best) the line passing through $A_1, A_2$ (quite generically actually it is just a point). So you cannot describe all lines with three colinear points.

*This last one is not a condition but a caution. In general $\sum s_i = 1$ and $\mathbf{n}\cdot \mathbf{s}=0$ does not have a solution, you must demand $\mathbf{n}$ to not be a multiple of $(1,1,1)$ to get solutions. So $L_{\mathbf{A},\mathbf{n}}$ can be empty! There is a one-to-one correspondence between lines in $\mathbb{R}^2$ and not-empty $L_{\mathbf{A},\mathbf{n}}$s.


Given these conditions let's show $L_{\mathbf{A}, \mathbf{n}}$ is a line. Take two points on $L_{\mathbf{A}, \mathbf{n}}$, say $P=\mathbf{p}\cdot \mathbf{A}$ and $Q=\mathbf{q}\cdot \mathbf{A}$. Both $\mathbf{p}, \mathbf{q}$ are vectors in $\mathbb{R}^3$ that live in the plane $\mathbf{s}\cdot \mathbf{n}=0$. Suppose $\mathbf{q}=\alpha \mathbf{p}$, then 
$$
1=\sum q_i = \alpha \sum p_i = \alpha
$$
So for two distinct points of $L_{\mathbf{A}, \mathbf{n}}$ the vectors $\mathbf{p}, \mathbf{q}$ are linearly independent. Now if they are linearly independent, since they both live in $\mathbf{s}\cdot \mathbf{n}=0$, their span is exactly the whole $\mathbf{s}\cdot \mathbf{n}=0$. Hence
$$
\mathbf{s} = t\mathbf{p} +r \mathbf{q}
$$
On the other hand if
$$
1=\sum s_i = t\sum p_i + r\sum q_i=t + r
$$
Therefore points on $L_{\mathbf{A}, \mathbf{n}}$ are exactly $t\mathbf{p}\cdot \mathbf{A}+(1-t)\mathbf{q}\cdot \mathbf{A}=tP+(1-t)Q$. This is a line passing through $P,Q$.
Conversely we need to show that any line $\ell$ and any three points, $\mathbf{A}$, that are not colinear, then one can find $\mathbf{n}$ such that $\ell = L_{\mathbf{A}, \mathbf{n}}$. This is done as follows: First note that the set
$$
X_{\mathbf{A}}= \{\mathbf{s}\cdot \mathbf{A}|\sum s_i =1\}
$$
is the whole $\mathbb{R}^2$ for any non-colinear $\mathbf{A}$. This is because if $A_1, A_2, A_3$ are not colinear, then the vectors $A_1-A_3, A_2-A_3$ are linearly independent, so the set of points
$$
s_1( A_1-A_3) + s_2(A_2-A_3)+A_3
$$
is the whole $\mathbb{R}^2$. Now given any line $\ell$, choose two points $P,Q$ on it. There exists $\mathbf{p},\mathbf{q}$ such that $\sum p_i = \sum q_i = 1$ and $\mathbf{p}\cdot \mathbf{A}=P$ and $\mathbf{q}\cdot \mathbf{A}=Q$, by what we just shown. The vectors $\mathbf{p},\mathbf{q}$ span a plane in $\mathbb{R}^3$, given by an equation $\mathbf{x}\cdot \mathbf{n}=0$ for some $\mathbf{n}$. But then any point $R(t)\in \ell$ is of the form
$$
R(t)=tP+(1-t)Q=[t\mathbf{p}+(1-t)\mathbf{q}]\cdot \mathbf{A}:=\mathbf{r}(t)\cdot \mathbf{A}
$$
Clearly $\sum r_i(t) = 1$ and $\mathbf{r}(t)\cdot \mathbf{n}=0$ for all $t$. Hence $\ell \subseteq L_{\mathbf{A}, \mathbf{n}}$. By what we proved previously this is actually an equality $\ell = L_{\mathbf{A}, \mathbf{n}}$. QED.
