Based on the answer given here, ($(A)!1cm!(B)$) in TikZ is equivalent to ([nodesep=1]{B}A) in PSTricks. It is just about the "radial" distance specified by two points. How about the "transverse" distance specified by two points?

More precisely, in PSTricks, the syntax ([offset=2]{A}B) represents a new point (namely C) which is 2 unit apart from the line joining A and B such that BC is perpendicular to AB.


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What is the TikZ equivalent for such a syntax?


([offset=2]{A}B) can also be written as ([nodesep=2,angle=90]{A}B). Other approaches are also possible with, for example, transformations provided by pst-eucl.


2 Answers 2


The equivalent syntax is ($(B)!2cm!90:(A)$), which is the point 2cm away from B on the line from B to A after that line has been rotated by 90 degrees.



\coordinate (A) at (1,1);
\coordinate (B) at (4,4);

\coordinate (C) at ($(B)!2cm!90:(A)$);
\node at (A) [above] {A};
\node at (B) [above] {B};
\node at (C) [above] {C};

\fill (A) circle [radius=2pt] (B) circle [radius=2pt] (C) circle [radius=2pt];

With tkz-euclide package, using syntax \tkzDefPointWith[orthogonal normed, K=2](B,A) \tkzGetPoint {C} define a point C from B perpendicular to AB with 2cm length.



 \tkzDefPoints{0/0/O, 5/5/K}
  \tkzDefPoints{1/1/A, 4/4/B}  
  \tkzDefPointWith[orthogonal normed, K=2](B,A) \tkzGetPoint {C}   
  \tkzDrawLine[color=blue,add=0 and 0](B,tkzPointResult)


enter image description here

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