Prove that two distinct points are contained by at most one line? Definition. A line is the set of points $(x,y)\in\mathbb{R}^2$ that satisfy the equation $$ax+by+c=0,$$ where at least one of $a$ or $b$ is non-zero.
How to prove that given two distinct points $(p,q)$ and $(r,s)$, there is at most one line that contains both points? (Edit: Ideally without much more knowledge than high school level algebra and arithmetic, so ideally without linear algebra.)
Unfortunately my proof (posted below) is long and inelegant, so I'm hoping for a short and elegant one.
 A: Well, the equation $ax=by$ describes all possible lines through the origin (also the vertical one). Now the line through $(p,q)$ can be described as
$$a(x-p)=b(y-q)$$
When does this include $(r,s)$? Well, if it does, we have
$$a(r-s)=b(s-q)$$
meaning $\tfrac{a}{b}=\tfrac{s-q}{r-s}$. Since $\tfrac ab$ describes the slope of the line, and $\tfrac ab$ is fixed by $p,q,r,s$ (and we've already got two points it has to pass through), we know there's only one. We do have to be a little careful as $r-s$ cannot be $0$, but the case where it is is easily handle-able (it means $b=0$ as $s-q\neq 0$ because the two points are distinct).
A: Let $(p,q)$ and $(r,s)$ be distinct points. Now, treat $a,b,c$ as variables and consider homogeneous linear system
$$ ap + bq + c = 0,\\ ar+bs+c = 0.$$
This system is of rank $2$, because vectors $(p,q,1)$ and $(r,s,1)$ are linearly independent if and only if $(p,q)\neq (r,s)$. This means that the solution space $L$ is one-dimensional. Pick any non-zero vector in $L$, say $(A,B,C)$. Notice that $A=B = 0$ would imply $C = 0$ as well, which is contradiction with our choice of $(A,B,C)$. Your line is then given by equation
$$Ax+By+C = 0.$$
Note that if we chose another non-zero $(A',B',C')$, then there would exist unique $\lambda\neq 0$ such that $(A',B',C') = \lambda(A,B,C)$ and thus
$$Ax+By+C = 0 \iff \lambda(Ax+By+C) = 0 \iff A'x +B'y + C' = 0$$ and this proves uniqueness.
EDIT: This was written in the language of linear algebra, but it can be easily modified to high school level. The basic idea is unchanged, you start with the system and simply solve it to get your line. The very fact that the equation you get is unique up to a scaling factor, proves uniqueness of the line.
A: My long and inelegant proof:
Suppose the lines $ax+by+c=0$ and $dx+ey+f=0$ contain both $\left(p,q\right)$ and $\left(r,s\right)$. Then:
$$ap+bq+c \overset{1}{=}0,
\\ar+bs+c \overset{2}{=}0,
\\dp+eq+f \overset{3}{=}0,
\\dr+es+f \overset{4}{=}0.$$
Taking $\overset{1}{=}-\overset{2}{=}$ and $\overset{3}{=}-\overset{4}{=}$, we have:
$$a\left(p-r\right) \overset{5}{=}b\left(s-q\right),
\\d\left(p-r\right) \overset{6}{=}e\left(s-q\right).$$
If $p=r$, then given that $\left(p,q\right)\neq\left(r,s\right)$, we have $s\neq q$. Since $p-r=0$ and $s-q\neq0$, $\overset{5}{=}$ and $\overset{6}{=}$ imply that $b=e=0$. That is, the coefficient on $y$ for each of our two lines is zero. Hence our two lines are simply $x=p$ and $x=r$. But of course, $p=r$ and so the two lines are identical.
Now suppose $p\neq r$. Then we have:
$$a \overset{7}{=}b\frac{s-q}{p-r},
\\d \overset{8}{=}e\frac{s-q}{p-r}.$$
Since at least one of $a$ or $b$ must be non-zero, $\overset{7}{=}$ implies that both $a$ and $b$ must be non-zero.
Similarly, since at least one of $d$ or $e$ must be non-zero, $\overset{8}{=}$ implies that both $d$ and $e$ must be non-zero.
Using $\overset{7}{=}$ and $\overset{8}{=}$, we rewrite $\overset{1}{=}$ and $\overset{3}{=}$ as:
$$\frac{s-q}{p-r}p+q+\frac{c}{b} \overset{9}{=}0,
\\\frac{s-q}{p-r}p+q+\frac{f}{e} \overset{10}{=}0.$$
Taking $\overset{10}{=}-\overset{9}{=}$, we have:
$$\frac{c}{b}\overset{11}{=}\frac{f}{e}.$$
We now show that a point $\left(t,u\right)$ is in the line $ax+by+c=0$ if and only if it is also in the line $dx+ey+f=0$. We will thus have shown that the two lines are identical:
\begin{alignat*}{2}
 &  & at+bu+c & =0\\
\overset{7}{\iff} &  & b\frac{s-q}{p-r}t+bu+c & =0\\
\iff &  & b\left(\frac{s-q}{p-r}t+u+\frac{c}{b}\right) & =0\\
\iff &  & \frac{s-q}{p-r}t+u+\frac{c}{b} & =0\\
\overset{11}{\iff} &  & \frac{s-q}{p-r}t+u+\frac{f}{e} & =0\\
\iff &  & e\left(\frac{s-q}{p-r}t+u+\frac{f}{e}\right) & =0\\
\iff &  & dt+eu+f & =0.
\end{alignat*}
A: Suppose that $Α(x_1,y_1)$ και $Β(x_2,y_2)$ belong to
\begin{align*}
\varepsilon_1&:a_1x+b_1y+c_1=0\\
\varepsilon_2&:a_2x+b_2y+c_2=0.
\end{align*}
Without loss of generality, let $y_1\neq y_2$.
Multiplying the first equality with $a_2$, the second equality with $a_1$ and subtracting the first equality from the second equality, we get the equation
\begin{equation}
(a_1b_2-a_2b_1)y+(a_1c_2-a_2c_1)=0.
\end{equation}
which has two different solutions $y_1$ και $y_2$. Since the last equation has a degree equal to one, this can only happen when
\begin{align*}
a_1b_2-a_2b_1&=0\\
a_1c_2-a_2c_1&=0
\end{align*}
or equivalently
\begin{equation*}
\begin{vmatrix}
a_1 & b_1  \\ 
a_2 & b_2
\end{vmatrix}
=
\begin{vmatrix}
a_1 & c_1  \\ 
a_2 & c_2
\end{vmatrix}
=0
\end{equation*}
This means that there exists $k \in \mathbb{R}$ such that
\begin{align*}
a_2&=ka_1\\
b_2&=kb_1\\
c_2&=kc_1\\
\end{align*}
Hence
\begin{align*}
\varepsilon_1&:a_1x+b_1y+c_1=0\\
\varepsilon_2&:ka_1x+kb_1y+kc_1=0.
\end{align*}
and so $\varepsilon_1$ and $\varepsilon_2$ are actually the same lines.
