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The geometric interpretation of the parametric equation of a line $\mathbf{w}+t\mathbf{u}$ is quite intuitive, you have a parameter that can expand, contract and change the direction of the line. How can I understand visually the cartesian equation of a line $ax + by = c$?

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  • $\begingroup$ as ax=c+(-b)y ? $\endgroup$
    – user645636
    Feb 3, 2020 at 21:51
  • $\begingroup$ Could you elaborate a bit more? $\endgroup$ Feb 5, 2020 at 0:16

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Try and rewrite the equation:

$ax + by = c$ as $y = (-\frac{a}{b})x + \frac{c}{b}$.

You can then simply see how $y$ changes as $x$ does (and see, for instance, that when $x = 0$, $y$ will take the value $\frac{c}{b}$, or that $y = 0$ when $x = \frac{c}{a}$. :)

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  • $\begingroup$ Thanks for answering but I was kind of trying to understand the formula as it is, because I supose there's a reason for it to be writen the way it is. $\endgroup$ Feb 3, 2020 at 22:22
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In Euclidean geometry (i.e with an inner product) take $(x_0,y_0)$ to be one point on the line, then $c=ax_0+by_0$ so the equation of the line rewrites as

$$a(x-x_0)+b(y-y_0)=0$$

And we can interpret that as

$$(a,b)\cdot \begin{pmatrix} x-x_0\\y-y_0 \end{pmatrix}=0$$

So the line passes through $(x_0,y_0)$ and has $(a,b)$ as normal vector

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  • $\begingroup$ Could you please explain how you got x - x0 and y - y0 inside the parenteses? $\endgroup$ Feb 3, 2020 at 22:24
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    $\begingroup$ $ax+by=c=ax_0+by_0\Rightarrow ax+by-ax_0-by_0=0 \Rightarrow a(x-x_0)+b(y-y_0)=0$ $\endgroup$
    – marwalix
    Feb 3, 2020 at 22:29
  • $\begingroup$ Just a couple more things. Why can I say that the line $c = ax_0 + by_0$ have the point $(x_0, y_0)$ on it? And because you have a line that goes through the point $(x_0, y_0)$ and have a $(a, b)$ as a normal vector, does this means that $a$ and $b$ are responsable for making the line not pass through the origin? Something like the linear component of a line in the form $y = ax + b?$ $\endgroup$ Feb 4, 2020 at 16:32
  • $\begingroup$ Because I chose $(x_0,y_0)$ as being one point on the line $\endgroup$
    – marwalix
    Feb 4, 2020 at 16:35
  • $\begingroup$ No because if $c=0$ the line $ax+by=0$ passes through the origin $\endgroup$
    – marwalix
    Feb 4, 2020 at 16:37
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Personally, I find it intuitive to visualize lines or planes by considering their gradients. One of the most useful aspects about the gradient of a function is that it tells us the direction of steepest ascent or increase of that given function.

To put this into perspective, let's consider your example. We can interpret $ax + by = c$ to be a particular instance of a general function $f(x, y) = ax + by$ where the $f$ evaluates to $c$. The gradient of $f$ can then be calculated as follows:

$$\nabla f = \left<\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right> = \left<a, b \right>$$

Because the gradient is a constant vector, we know that the function's direction of increase on a contour plot is going to be the same at any arbitrary point $(x, y)$ no matter where you evaluate the function. From this, we can intuit that the function is linear and perpendicular to the vector $\left<a, b\right>$.

Note that we can extend this idea to any number of higher dimensions. For example, we know that $f(x, y, z) = ax + by + cz$ is a plane that is orthogonal to the vector $\left<a, b, c \right>$. Such a vector is often referred to as the normal vector of the plane.

If anything is unclear, I'd be more than happy to elaborate.

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  • $\begingroup$ Thanks for answering but I don't feel very confortable with gradients to understand your answer completely. $\endgroup$ Feb 4, 2020 at 17:01
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It helps let you look at a circle as a line just substitute $a=x$, $b=y$ and you've got a circle around the origin with radius $\sqrt{c}$ . in general $a$ is the scaling of $x$, $b$ the scaling of $y$ . Scale each side by a factor, it's the same line. Negative $b$ to reflect across the $x$ axis, negative $a$ opposite slope through same point. negate both to negate the intercept , as will negating $c$ . make $a$ change magnitude, and it rotates around a point. etc.

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  • $\begingroup$ That's a good way to see it and scaling $x$ and $y$ was the first approach that I had, but I couldn't figure out where the $c$ in the equation comes from. Maybe I should read a book on analityc geometry? $\endgroup$ Feb 5, 2020 at 17:20
  • $\begingroup$ I don't know that topic myself. $c$ is just a numerator in the intercept to me. $\endgroup$
    – user645636
    Feb 5, 2020 at 17:22

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