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First I will determine that given any two points there exists a unique line through them. Let $(x_1,y_1)$ and $(x_2,y_2$ be two distinct points. We know that given any two points there exists a line containing both points. Now let $ax+by=c$ be an equation of a line containing both points. Since the equation contains both points we have that

$ax_1+by_1=c$ Equation $1$

$ax_2+by_2=c$ Equation $2$

Now consider two cases:

Case $1$: $x_1=x_2$

Subtracting equation $2$ from $1$ we get that $b(y_2-y_1)=0$. Now since $(x_1,y_1) \not =(x_2,y_2)$ and $x_1=x_2$ we have that $y_1\not =y_2$. Hence it follows that $y_2-y_1 \not =0$. So we have that $b=0$, hence $a\not =0$ or else $ax+by=c$ would not be a line. Thus the equation $ax+by=c$ is equivalent to $x=\frac{c}{a}$. Then it follows from equation $1$ that $ax_1=c$ and so $\frac{c}{a}=x_1$. Hence the equation $ax+by=c$ is equivalent to the equation $x=x_1$ which is unique.

Case $2$: $x_1\not =x_2$

Suppose that $b=0$. It follows from equation $1$ and $2$ that $ax_1=c$ and $ax_2=c$ thus $ax_1=ax_2$. So $a(x_1-x_2)=0$. However $x_1\not =x_2$ and thus $x_1-x_2 \not =0$. Thus $a=0$ so we have that $(a,b)=(0,0)$ which is a contradiction. Thus it must be that $b \not =0$, then the equation $ax+by=c$ becomes equivalent to $y= -\frac{a}{b}x + \frac{c}{b}$. Now subtracting equation $1$ from $2$ we get that $ax_2-ax_1+by_2-by_1=0$. Thus $b(y_2-y_1)=-a(x_2-x_1)$, since $x_2-x_1 \not =0$ we have that $-\frac{a}{b}= \frac{y_2-y_1}{x_2-x_1}$. Dividing equation $1$ by $b$ we have that:

$$\frac{c}{b}=\frac{a}{b}x_1+y_1=-\frac{y_2-y_1}{x_2-x_1}x_1+y_1=\frac{x_2y_1-x_1y_2}{x_2-x_1}$$

Hence the equation $ax+by=c$ is equivalent to the equation $y=\frac{y_2-y_1}{x_2-x_1}x+\frac{x_2y_1-x_1y_2}{x_2-x_1}$ which is unique.

Now given $3$ non colinear points $A,B,C$ we have 3 unique lines $\overleftrightarrow{AB}$, $\overleftrightarrow{AC}$,$\overleftrightarrow{BC}$.

Looking for suggestions on alternative shorter cleaner proofs or if this is an ok approach.

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It depends a lot on what axioms and definitions you use.

So there are 3 points. For every pair of points there exists at least a line connecting them. Actually it's exactly one line if the points are distinct, but usually infinitely many lines if the points are the same. The fact that two distinct points always define a line is often an axiom in a formal definition of planar geometry. So I'd only take the algebraic path you took if I had some specific algebraic definitions of what a point or a line is. Such definitions would need to address the fact that the same line can be described by different equations, namely multiples of one another, which easily leads to concepts like equivalence classes and homogeneous coordinates.

Now you have that there has to be a line for every pair of points. If two of these lines were the same, then the three points involved would be collinear. By problem statement they are not, so there have to be at least three lines. Note that this also covers the situation where two of the points are the same: in that case any line incident with one of them will be incident with the other as well, so by definition your points would be collinear as well. You can only ever have non-collinear points if they are distinct.

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