So I found: $ax+by=c$
What exactly does $c$ stand for? What does it do and why is $c$ important? I googled everywhere but couldn't find an answer other than $c$ being a constant.
So I found: $ax+by=c$
What exactly does $c$ stand for? What does it do and why is $c$ important? I googled everywhere but couldn't find an answer other than $c$ being a constant.
On
If $b=0$, then $$ ax+by=c \iff ax=c $$
which means $x$ is a constant.
Otherwise, $$ y= c/b-(a/b) x $$
That is $c/b$ is the initial value of $y$ associated to the $x=0$, that is the $y$ -intercept.
On
There is a straight explanation for it. First write the equation in slope intercept form:-
$$y=-Kx+J$$ I am using $K$ in place of slope and $J$ for y intercept.( Note I used -K so that you get the desire form orstill you would get the standard for! But for the sake of simplicity I would use -K)
Now :-
$$y+Kx=J \\ \implies Kx+y=J$$
Slope is generally a fraction if it is a fraction you have B with y or else your B would be 1 in standard form. But let us assume that $K$ is a fraction equal to $\frac{A}{B}$
Then :-
$$\frac{A}{B}x+y=J$$ I would multiply both sides by B.
$$B(\frac{A}{B}+y)=BJ$$ I would simplify it directly in the next equation so please see it.
Let $BJ = C$
Then by distribution of B and the above assumption:-
$$Ax+By=C$$
And here is our Answer. If I took K positive slope then it would have been -Ax instead try yourself.
Since you filed this under Linear Algebra, I will take an LA approach towards it.
Your equation is the 2D analog of the 3D equation for a plane defined by its normal. In 3D, a point can be represented as a vector $ \vec p = (x,y,z)$. Another vector is $(a,b,c)$. Suppose $(a,b,c)$ is the normal $\vec n$ to a plane passing through a specific point $ \vec p_0 = (x_0,y_0,z_0)$. Any point on the plane has to have the condition the the vector formed by it and $\vec p_0$ has to be perpendicular to $\vec n$:
$$ ( \vec p - \vec p_0 ) \cdot \vec n = 0 $$
$$ \vec p \cdot \vec n = \vec p_0 \cdot \vec n $$
$$ a x + by + cz = a x_0 + by_0 + cz_0 = d $$
Because $\vec n$ and $\vec p_0$ don't vary, the value of $d$ is constant. The value determines where along $\vec n$ the plane is actually located.
In the 2D case, the normal to the line is $(a,b)$ and $c$ plays the role of $d$. In other words, $(a,b)$ tell the direction of the line and $c$ gives you the location.