Why does all points $(x,y)$ satisfying $ax+by=c$ stay on a straight line?

Question

We know that points $(x_i,y_i)$ which satisfy the equation $ax_i+by_i=c$ lie on the same straight line. I understand that all points on this line satisfy the equation, but how do we ensure that all points satisfying the equation stay on the same line?

As @saulspatz says,

How do we know that the lines defined by analytic geometry satisfy the Euclidean axioms?

If $a,b$ are not both zero, then $ax+by=c$ is the equation of a line. This means: the set of all solutions is the line. — GEdgar, Apr 25 '21 at 12:37
As GEdgar said, one way to think about it is that it is a line by definition. If you want something else, you should first tell us what do you mean by line, i.e. give us definition you want to consider. — Ennar, Apr 25 '21 at 12:40
@ennar Why does all points on the line satisfy the equation? — Eisenstein, Apr 25 '21 at 12:45
Am I right in thinking that you are asking how we know that the lines defined by analytic geometry satisfy the Euclidean axioms? — saulspatz, Apr 25 '21 at 12:45

score 1 · Accepted Answer · answered Apr 25 '21 at 12:49

$$ax+by=c$$

If $a \ne 0$, then we have $x =\frac{c-by}a$. Knowing $y$ completely determines $x$. Let $y$ takes value $t$

$$\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} \frac{c}{a}\\0\end{bmatrix} + t\begin{bmatrix} -\frac{b}{a}\\ 1\end{bmatrix}$$

The locus is parallel to the direction of $\begin{bmatrix} -\frac{b}{a} \\ 1\end{bmatrix}$ and passes through the point $\begin{bmatrix} \frac{c}{a}\\0\end{bmatrix} $. Since it moves along a direction, it is a line.

If $a=0$ and $b\ne 0$, then $y=\frac{c}{b}$ and $x$ can takes any values, this is clearly a horizontal line.

Why does all points $(x,y)$ satisfying $ax+by=c$ stay on a straight line?

1 Answers1