# Given $3$ non co-linear points in a plane there are at least $3$ lines.

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.