The equation of a line is of the form:
\begin{align*} Ax + By + C = 0 \end{align*}
My question is, how has this equation been obtained? Why, for example, does the term $A$ multiply $x$, instead of dividing it or simply not being? Why does this equation describe all those points and what is the proof that this is true?
Thanks in advance.
The general form $Ax+By+C=0$ is being used to describe all possible lines in $2D$. Why do we multiply by $A$? Because we cannot divide by zero. Multiplication avoids such issues and allows us to describe lines such as $x=3$ or $y=-1$. Why the graph of all solutions to a linear equation is a line? Think that line has always the same slope at any point and this slope is defined by an angle between the line and axis $X$. That angle can be described by a tangent which is equal to the ratio of change in $y$ to change in $x$. If you have two points $(x_1,y_1)$ and $(x_2,y_2)$ which are solutions of $Ax+By+C=0$ then we have $Ax_1+By_1+C=0$ and $Ax_2+By_2+C=0$. The tangent ratio (slope) will be $\frac{y_2-y_1}{x_2-x_1}=\frac{C-Ax_2-C+Ax_1}{B(x_2-x_1)}=-\frac{A}{B}$. As you can see, the slope does not depend on $x_1, x_2, y_1, y_2$ which means it will be the same for any pair of points.
The equation of a line is obtained from a direction and a point. Without loss of generality, let us assume that $\textbf{d} = (d_{1},d_{2})$ is the direction where $d_{1}d_{2}\neq 0$ and $\textbf{P} = (p_{1},p_{2})$ is some point of it. Thus the points $(x,y)$ of the line passing through $\textbf{P}$ along the direction $\textbf{d}$ must satisfy: \begin{align*} & (x,y) = \lambda\textbf{d} + \textbf{P} = \lambda(d_{1},d_{2}) + (p_{1},p_{2}) = (\lambda d_{1} + p_{1},\lambda d_{2} + p_{2}) \Longleftrightarrow\\\\ &\begin{cases} x = \lambda d_{1} + p_{1}\\ y = \lambda d_{2} + p_{2} \end{cases} \Longleftrightarrow \lambda = \frac{x-p_{1}}{d_{1}} = \frac{y - p_{2}}{d_{2}} \Longleftrightarrow d_{2}x - d_{1}y - p_{1}d_{2} + d_{1}p_{2} = 0 \\\\ &\therefore Ax + By + C = 0\,\,\text{where}\,\,A = d_{2},\, B = -d_{1}\,\,\text{and}\,\,C = -p_{1}d_{2} + d_{1}p_{2} \end{align*}
If $d_{1} = 0$, you obtain a vertical line. If $d_{2} = 0$, you obtain a horizontal line.
"how has this equation been obtained?"
$y = mx + b \iff -mx + y - b = 0 \iff -mkx + ky - bk = 0; k \ne 0 \iff Ax + By + C=0; A=-mk; B=k; C =-bk; k\ne 0$.
....
$Ax + By + C = 0$ doesn't "mean" anything.
It's just that:
Case 1: $B \ne 0$.
$Ax + By + C = 0 \iff$
$By = -Ax - C \iff$
$y = -\frac ABx -\frac CB \iff$
$y = mx + b; m = -\frac AB; b = -\frac CB$.
Now the equation $y = mx + b$ for a line DOES mean something.
$m$ is the slope (rise/run) of a line. By the geometric nature of a line (as opposed to a curve) we know this always exists and is constant. And $b$ is the $y$ value where the line crosses the $y$-axis. (This always occurs unless the line is straight up and down vertical.)
Case 2: $B = 0; A\ne 0$.
Then $Ax + By + C = 0\implies Ax = -C \implies x = -\frac CA$.
If we set $d = -\frac CA$ then this is a straigh up and down vertical line where $x = d$.
That's all.
Case 3: $A=0; b=0$
The this is just a statement $C = 0$. It's not the eqaution of a line. It's just a statement that there is a constant $C = 0$.
....
So $Ax + By + C = 0$ will always be a line (unless $A = B = 0$). Not because $A,B,C$ mean anything but because they can be manipulated into something that does.
===== old answer below... which explains why $y = mx + b$ is the equation of a line ====
A point of a line is $(c,d)$ and if you travel along the $x$ axis some distance of $t$ then you must travel up the $y$ axis for some proportional distance of $m*t$ where $m$ is the slope of this particular line.
So you end up at the point $(c + t, d + m*t)$.
Now $t$ can be any distance. So for any variable of $t$ distance you have $(x,y) = (c + t, d+m*t)$ with $x = c+t$ and $y = d + m*t$.
If we express this in terms of $x$ (rather than in terms of $t$ we get)
$x = c + t$ so $t = x - c$ so $y = d + m *t = d + m(x - c) = mx + (c-md)$.
So the equation of a line is $y = mx + (c-md)$.
This is interpreted as $m =$ the slope of the line. And $c$ can be the $x$ coordinate of any point on the line and $d$ is the coresponding $y$ coordinate.
If we plug in $x=0$ we get $y = c-md = b$ and this is the $y$-intercept of the line (the $y$ coordinate when the line crosses the $y$-axis.)
This gives us the more common equation of the line $y = mx + b$.
We can convert this to your form by multiplying both sides by any non-zero $B$. Then we get: $By = mBx + bB$
$-mBx + By -bB = 0$.
$Ax + By + C = 0$ where $A = -mB$ and $C = -bB$
The numbers $A, B, $ and $C$ have no meaning or significance by the do have a relationship that $-\frac AB = m$ the slope of the line. And $-\frac CB$ is the $y$ intercept of the line and $-\frac CA$ is the $x$ intercept of the line.
In other words:
If you have an equation $Ax + By + C = 0$ you know it is a line.
You know that if you plug in $x=0$ you get $y = -\frac CB = b$ is the $y$-intercept. And you know that $y = -\frac ABx - \frac CB$ so that for every unit of $1$ that $x$ changes then $y$ will change by $-\frac AB$ so the $-\frac AB = m$ is the slope of the line.
