Is there an easy way to store the Euclidian distance between two coordinates?

I want to store the value 4 in a macro, so that I can use it to specify the width of a node.

The easiest way I could think of to to this is to somehow store the output of veclen(x,y) to a macro. I found this line in the tikz-pgf manual

\pgfmathparse{veclen(12,5)} \pgfmathresult

but I can't get it to work.

Does anyone have an idea? I found a couple of answers in this site that have something to do with calculating the distance between coordinates, but they always use the distance inside a specific path and don't store it in a macro to use it later.

3 Answers 3


Here is a simple (?) code that computes the distance and stores it in a variable.

\newcommand{\Distance}[3]{% % from https://tex.stackexchange.com/q/56353/121799
}% Explanation: the calc library allows us, among other things, to add and
% subtract points, so ($#1-#2$) is simply the difference between the points
% #1 and #2. The combination \tikz@scan@one@point\pgfutil@firstofone extracts
% the coordinates of the new point and stores them in \pgf@x and \pgf@y. 
% They get fed in veclen, and \pgfmathsetmacro stores the result in #3. 
% EDIT: included fudge factor, see https://tex.stackexchange.com/a/22702/121799
\coordinate (X) at (1,0);
\coordinate (Y) at (3,0);
\draw[-] (X) -- (Y);
\node at (0,2) {\Distance{(X)}{(Y)}{\mylen}\mylen};
  • Thanks, that works! I'll try to understand your code which may take me a while since I'm a beginner.
    – Florian
    Jan 30, 2018 at 18:06
  • @Florian I added an explanation.
    – user121799
    Jan 30, 2018 at 18:17
  • Thanks for adding the explaination. Is there a way to eliminate the rounding(?) error? veclen(1,-4) yields 4.12305, and your macro 4.13846. I dont' think the small difference matters for my use of the macro, I'm just curious.
    – Florian
    Jan 30, 2018 at 18:23
  • @Florian Yes, see the update.
    – user121799
    Jan 30, 2018 at 18:31
  • 1
    @PiFisher I am not sure if I understand your question, but internally TikZ is using points. \tikz@scan@one@point\pgfutil@firstofone gets the point (here defined as the difference between two points) and stores its coordinates in \pgf@x and \pgf@y, and the units used there are points. However, the OP wanted cm, that's why I convert.
    – user121799
    Feb 28, 2018 at 4:19

Either calculate it manually using sqrt(x^2 + y^2) or use veclen as suggested:

enter image description here




% Euclidian distance from (3, 4) to (12, 5)
\pgfmathparse{sqrt((12 - 3)^2 + (5 - 4)^2)}\pgfmathresult

% Euclidian distance from (3, 4) to (12, 5)
\pgfmathparse{veclen(12 - 3, 5 - 4)}\pgfmathresult


There is a rounding difference.


For some particular cases you can also use a fit node. This way there's no need for storing the distance value.

\usetikzlibrary{positioning, fit}


\draw[fill=black] circle (1pt) node[below]{(0,0)} -- node[above]{4} ++(4,0) circle (1pt) node[below] {(4,0)};

\node[fit={(0,0) (4,0)}, inner xsep=0pt, draw, minimum height=1cm, anchor=south west] at (0,5mm) (a) {};

enter image description here

  • Thanks, I didn't know there is such an option.
    – Florian
    Jan 30, 2018 at 18:28
  • @Florian, latex is full of options :)
    – CroCo
    Jan 31, 2018 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.