How to prove that through two distinct points, there passes a unique line? I start with analytic geometry approach, i.e. two distinct points $P$=$(x1,y1)$ and $Q$=$(x2,y2)$ with case $x1=x2$ and conclude that the equation of the line is actually $x=x1$ meaning that the value of $x$ is unique. Is my proof true for this case?
Meanwhile, I am stuck with the second possibility, when $x1$ is not the same as $x2$
Please your help
 A: If your definition of a line in $\mathbb{R}^2$ is any set of the form
$$
L_{m,b}:=\{(x,y)\in\mathbb{R}^2:y=mx+b\}
$$
or
$$
V_c:=\{(x,y)\in\mathbb{R}^2:x=c\}
$$
(vertical lines)
then you can proceed like this.
Case 1: Suppose $x_1=x_2$.
Clearly, the only vertical line passing through $P$ and $Q$ is $V_{x_1}$.
Now, no line of the form $L_{m,b}$ passes through $P$ and $Q$. To show this, suppose that
$$
y_1=mx_1+b\\
y_2=mx_2+b
$$
Then,
$$
b=y_1-mx_1=y_2-mx_2\tag{1}
$$
so that
$$
y_2-y_1=m\underbrace{(x_2-x_1)}_{0}=0\tag{2}
$$
and $y_1=y_2$. This contradicts the assumption that $P\neq Q$.
Case 2: Suppose $x_1\neq x_2$.
Clearly no vertical line passes through $P$ and $Q$.
Suppose that $L_{m,b}$ passes through $P$ and $Q$. Then, as in $(1)$, we must have
$$
m=\frac{y_2-y_1}{x_2-x_1}\tag{3}
$$
and from $(1)$ we must have
$$
b=\frac{y_1x_2-y_2x_1}{x_2-x_1}\tag{4}
$$
This shows unicity of $m$ and $b$ and since it is readily verified that the line $L_{m,b}$ with $m$ and $b$ as in $(3)$ and $(4)$ passes through $P$ and $Q$, we have existence.
