# Talk:Distance from a point to a line

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## Algebraic proof

There is a major jump in the algebraic proof when it begins with "Then it is necessary to show.."

I agree. I rephrased the vector version to be more detailed. I'm not going to touch the algebraic version -- it's way more complicated and has more ugly details than the vector version. I added reqdiagram because a diagram would really help the vector explanation. —Ben FrantzDale (talk) 12:32, 17 August 2011 (UTC)

## Images

We would like to add images to this page, but because we are new users we are not allowed to upload files. If anyone would like to assist, we found some images at [1] that we believe would be helpful. Thanks! Kcoccio024 (talk) 18:32, 6 December 2013 (UTC)

## Sphere

Ah, but the surface of the earth is more like a sphere. Mention how to deal with that too. Jidanni (talk) 12:23, 22 December 2013 (UTC)

## Possible mathematical error?

The very first section of this page, titled Cartesian Coordinates appears to be wrong. I spent a good while being confused as to why a mathematical computer program I was writing was malfunctioning, until I realized that the following equation (which I was trying to use) doesn't seem to be true at all:

${\displaystyle \operatorname {distance} (ax+by+c=0,(x_{0},y_{0}))={\frac {|ax_{0}+by_{0}+c|}{\sqrt {a^{2}+b^{2}}}}.}$

Consider the very basic line, y = x.

The standard form of this equation (ax + by + c = 0) is: -x + y = 0. This means that:

a = -1 b = 1 c = 0

These values satisfy the conditions listed on the article: "where a, b and c are real constants with a and b not both zero"

Yet clearly, the distance equation listed will always return a distance of zero for any point.

Alexanderzero (talk) 06:16, 13 January 2014 (UTC) alexanderzero

I don't see why you think you only get zero distances (you do if you only deal with points on the line y = x). Try the point (1,2). It's distance from y = x is 1/√2.Bill Cherowitzo (talk) 04:31, 14 January 2014 (UTC)
As Alexanderzero sayed, the equation is wrong. It returns only the correct distances for horizontal, vertical and diagonal (with an angle of 90°)lines. As Alexanderzero, I spend some time to figure that out. To save other people's time, I would suggest to correct it. One correct equation for calculating the distance can be found here: http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.htmlJanDotNet (talk) 06:35, 8 June 2017 (UTC)
I am puzzled by your response. Look at equation (11) of your source. It is identical to the formula used on this page.--Bill Cherowitzo (talk) 15:29, 8 June 2017 (UTC)

### Section "Vector formulation" is also wrong

My computations show that the formula in Section "Vector formulation" is also wrong. For a correct formula (written in details for the 3d case, but siutable for n dimensions as well), see http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html. ${\displaystyle d^{2}}$ is shown in (6). — Preceding unsigned comment added by Makrai (talkcontribs) 10:15, 13 March 2014 (UTC)

I would check your computations again. The formula derived in the article is a simple application of vector projection and is not in error. If you think you have a counter-example please provide the details here. (The supposed error of the previous post did not exist, so your use of "also" is not quite correct.)Bill Cherowitzo (talk) 22:54, 13 March 2014 (UTC)

## Geometric proof???

The statements under the heading of Proof 2 (geometric proof) do not form a proof (the unjustified statement about the ratio of the sides of the right triangle requires a proof and has exceptions if either a or b is 0). Nor is this argument particularly geometric - the coordinate computations are just not presented. I consider this section just a piece of incorrect OR and propose that we get rid of it and replace it with a proof based on geometric transformations (say a well chosen rotation about the given point). Unfortunately I don't have a ready reference for such a proof, does anyone know of one? Bill Cherowitzo (talk) 05:05, 15 January 2014 (UTC)

I found a correct geometric proof (using similar triangles) and have replaced the suspect one. I still think that a transformation proof would be a nice addition. Bill Cherowitzo (talk) 22:59, 20 January 2014 (UTC)

## Last section

The recent edit that placed the two point version of the formula into the Cartesian coordinate section, while not a bad edit, has created a problem with the last section of this article. That section is devoted to this version of the formula and so is now redundant. There is some additional material in this section and my question is – is any of it worth saving? Bill Cherowitzo (talk) 19:13, 7 December 2014 (UTC)

I've deleted the now-redundant section, preserving the additional material which I've put into the Cartesian coordinate section. I've done a bot of fiddling to make it all a coherent whole. I hope this works for everyone. DOwenWilliams (talk) 22:53, 8 December 2014 (UTC)

## Even easier way for Vector formulation, incl. signed distance

It's trivial to create a Vector orthogonal to n (which, as n is supposed to be a unit vector, is one as well): ${\displaystyle o=(n.y,-n.x)}$

It's n rotated 90° clockwise.

Now one can just project the vector between a and p onto this orthogonal vector: ${\displaystyle signedDistance(x=a+tn,p)=(p-a)*o}$

It will be a positive value if it's on the right side of the line (relative to n), negative if it's on the left side.

If you only want the distance without a sign, just its absolute value. — Preceding unsigned comment added by 31.18.153.90 (talk) 01:55, 15 February 2015 (UTC)

The problem is that your 'trivial' first step is restricted to two dimensions; it does not work in higher dimensions. Given that, you've not really got a vector method but a 2D coordinate method, which must get the same answer as the earlier 2D formula.--JohnBlackburnewordsdeeds 02:28, 15 February 2015 (UTC)
The current formula assumes a "1x2 vector", which is 2D as well. But it mentions it also works for more than two dimensions so yeah, this simple way wouldn't work there. And, thinking about it, it only works for cartesian coordinate systems. Still, it's a nice and simple way for a signed distance in the common cartesian 2D case. — Preceding unsigned comment added by 31.18.153.90 (talk) 02:44, 15 February 2015 (UTC)

## Nomenclature in "Vector formulation"

The nomenclature in the "Vector formulation" section is inconsistent/ambiguous. The same labels are being used for points and vectors, which will confuse readers. For example, vector p might describe the location of point P with respect to the origin.
(Incidentally, I prefer to stick to the NIST/IUPAC/ISO standard [1][2][3] to set all variables in italic, including vectors.)
—DIV (120.19.123.255 (talk) 13:52, 30 August 2016 (UTC))

References

1. ^ Mills, I. M.; Metanomski, W. V. (December 1999), On the use of italic and roman fonts for symbols in scientific text (PDF), IUPAC Interdivisional Committee on Nomenclature and Symbols, retrieved 9 November 2012.
2. ^ See also Typefaces for Symbols in Scientific Manuscripts, NIST, January 1998. This cites the family of ISO standards 31-0:1992 to 31-13:1992.
3. ^ "More on Printing and Using Symbols and Numbers in Scientific and Technical Documents". Chapter 10 of NIST Special Publication 811 (SP 811): Guide for the Use of the International System of Units (SI). 2008 Edition, by Ambler Thompson and Barry N. Taylor. National Institute of Standards and Technology, Gaithersburg, MD, U.S.A.. March 2008. 76 pages. This cites the ISO standards 31-0:1992 and 31-11:1992.

## Improper use of the word "distance"

From the geometrical point of view it makes no sense to say "shortest distance" because by definition there is only one distance. If it's not "the shortest", it's not a distance. It would be better to say: "the shortest length among the length of the segments from the point and any point of the line". OK, I wrote it in a very long way, it could be shorter with somenthing implied. --Angelo Mascaro (talk) 15:22, 30 November 2016 (UTC)

While you are making a valid point, you are fighting an uphill battle. The English language does not always provide the exactness that a mathematician requires. In this case, the phrase "the shortest distance from a point to a line", has always stood as an abbreviation of – the shortest, among all the distances from a fixed point to any point on a line. Distance, in its most basic form, is defined between points. Using this basis the concept of distance can then be extended, as it is here, to distances between points and sets of points. Although the word is the same, this is a higher order concept of distance. The lead sentence (before I changed it) was not a definition as it simply repeated the phrase. What my edit did was to turn the lead into a definition by defining the concept in terms of the more primitive distance between points. This could all be done in terms of lengths of line segments as you have suggested, but I am not sure that that approach is clearer.--Bill Cherowitzo (talk) 19:11, 30 November 2016 (UTC)