# What is the TikZ equivalent for ([offset=2]{A}B)?

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.

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl}
\begin{document}
\begin{pspicture}(6,6)
\pstGeonode(1,1){A}(4,5){B}
\pcline[nodesep=-1](A)(B)
\pstGeonode([offset=2]{A}B){C}
\psline[linecolor=blue](B)(C)
\end{pspicture}
\end{document}


What is the TikZ equivalent for such a syntax?

## Miscellaneous

([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.

-

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.

\documentclass[tikz]{standalone}

\usetikzlibrary{calc}
\begin{document}

\begin{tikzpicture}
\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};

\end{tikzpicture}
\end{document}

-

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.

\documentclass{standalone}

\usepackage{tikz,tkz-euclide}
\usetkzobj{all}
\begin{document}

\begin{tikzpicture}
\tkzInit
\tkzDefPoints{0/0/O, 5/5/K}
\tkzDefPoints{1/1/A, 4/4/B}
\tkzDrawPoint[color=red](A)
\tkzDrawPoint[color=red](B)
\tkzLabelPoints[above](A,B)
\tkzDefPointWith[orthogonal normed, K=2](B,A) \tkzGetPoint {C}