# What is the geometric interpretation of a cartesian equation of a line?

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 $$ax + by = c$$?

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

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 $$x = \frac{c}{a}$$. :)

• 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. Feb 3, 2020 at 22:22

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

• Could you please explain how you got x - x0 and y - y0 inside the parenteses? Feb 3, 2020 at 22:24
• $ax+by=c=ax_0+by_0\Rightarrow ax+by-ax_0-by_0=0 \Rightarrow a(x-x_0)+b(y-y_0)=0$ Feb 3, 2020 at 22:29
• 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 $y = ax + b?$ Feb 4, 2020 at 16:32
• Because I chose $(x_0,y_0)$ as being one point on the line Feb 4, 2020 at 16:35
• No because if $c=0$ the line $ax+by=0$ passes through the origin Feb 4, 2020 at 16:37

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:

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

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 $$\left$$.

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 $$\left$$. 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.

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.
• 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? Feb 5, 2020 at 17:20
• I don't know that topic myself. $c$ is just a numerator in the intercept to me.