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I'm well used to the point-slope form and the $y$-intercept form of straight lines on a Cartesian plane, but I often see lines described as $ax + by + c = 0$. It is not easy for me to picture this the graph of the line this way.

Question. So what's so special about this form? When would a mathematician prefer this form over the others?

I've been trying to build an intuition for how to read the general form. It doesn't seem very intuitive. I think I have to say something like --- walk $b$ units along the $x$-axis and go up (or down) $a$ units and finally move $c$ units vertically. (But I must worry about whether both $a, b$ are negative, which gives me four possibilities. It's not convenient. There must be something interesting about the general form?)

user1145880
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    The preponderance of this form over $y=mx+c$ has always baffled me, similar to decimal places vs significant figures. The only benefit I can see is you can see such a curve is a line, and this also handles vertical lines. I thought before it useful for distance between a point and a line, but with $y=mx+c$ it's possibly easier. – JP McCarthy Apr 10 '23 at 02:14
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    The vector $(a,b)$ is perpendicular (aka, "normal") to the line. This generalizes nicely to "flat" objects in higher dimensions: the vector $(a,b,c)$ is perpendicular to the plane $ax+by+cz+d=0$, etc. You can also think of $ax+by+c$ as a function of $(x,y)$ that evaluates to zero for points on the line, positive for points on one side, and negative for points on the other side. Even better, scaling so that $a^2+b^2=1$, that function gives signed distance from the line. (This also generalizes to higher dimensions.) The form also treats $x$ and $y$ as "peers", which seems fair. – Blue Apr 10 '23 at 02:15
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    That said, the "preferred" form of a line (or any object) depends upon context. Point-slope form, standard form, parametric form, etc, all have their uses, advantages, and drawbacks. (For instance, point-slope form doesn't like vertical lines; standard and parametric forms don't mind them at all.) – Blue Apr 10 '23 at 02:22
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    This old answer of mine may be useful. The question is about the intuitive meaning of the ellipse equation, but I discuss forms of line equations. – Blue Apr 10 '23 at 02:29
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    @ryang: Done! – Blue Apr 10 '23 at 02:53
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    More on the normal vector $(a,b)$ view: when moving the constant to RHS as $ax+by=-c$, the LHS is the dot product between $(a,b)$ and $(x,y)$. The equation means all points on the line have the same scalar projection onto the normal vector. – peterwhy Apr 10 '23 at 03:08
  • When it gets to systems of linear equations "standard form" is generally easier to work with than slope-intercept or point-slope. Each form has its advantages. For example, slope-intercept is easy to graph. – user317176 Apr 11 '23 at 00:17

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  • The vector $(a,b)$ is perpendicular (aka, "normal") to the line. This generalizes nicely to "flat" objects in higher dimensions: the vector $(a,b,c)$ is perpendicular to the plane $ax+by+cz+d=0$, etc.

  • You can think of $ax+by+c$ as a function of $(x,y)$ that evaluates to zero for points on the line, positive for points on one side, and negative for points on the other side. Even better, scaling so that $a^2+b^2=1$, that function gives the exact signed distance from a point to the line. (This also generalizes to higher dimensions.)

  • The form also treats $x$ and $y$ as "peers", which seems fair. :)

That said, the "preferred" form of a line (or any object) depends upon context. Point-slope form, standard form, parametric form, etc, all have their uses, advantages, and drawbacks. (For instance, point-slope form doesn't like vertical lines; standard and parametric forms don't mind them at all.)

See also some thoughts in this old answer of mine. The topic is ellipse equations, but I start by discussing lines; and the general point about the benefit of flexibility applies.

Blue
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  • Treating $x$ and $y$ as peers is actually pretty cool. (My favorite property now.) Thanks very much for your attention. This is a very nice answer. You're saying that it's easy to picture the line in the standard form --- the normal vector gives me the inclination. (I don't yet see how to easily get the $y$ intercept, but I have only done a few seconds of pondering on this. I will most likely find a nice way out.) Thank you! – user1145880 Apr 10 '23 at 23:33
  • @user1145880: Happy to help! :) ... "You're saying that it's easy to picture the line in the standard form" Not really. As I mention in my old answer, different forms of an eqn tend to highlight different aspects of the geometry; standard forms typically indicate that the geometry is unimportant (compared to, say, the algebra of polynomials and whatnot). That we see the normal vector so easily is almost accidental, since standard form happens to be so very close to normal form $ax+by=d$. Standard form isn't trying to show you any geometry. – Blue Apr 11 '23 at 14:16
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$y=mx+d$ doesn't allow for vertical lines, but $ax+by+c=0$ does.