TikZ: how to find the intersection of two extended lines?

Question

From How to draw a line passing through a point and perpendicular to another? I learned to use shorten to extend lines. But the extended part doesn't seem to "really exist". For example, I would like to connect two predefined dots and extend the line segment by 1cm and then define the new end as "C" for later use. But \draw[shorten >=-1cm] (A) -- (B) coordinate (C) will only give me a C which is actually still B.

What I really need to do is to find the intersection of two extended lines, but I think the tough part is to make the extended lines "real".

Answer 1

You're right, the shorten > syntax doesn't "actually" extend the line, but works on a lower level. In your case, I would use \usetikzlibrary{calc}, which gives you access to a lot of nifty coordinate calculations.

Your desired result could be achieved with the syntax

\draw (A) -- ($(B)!-1cm!(A)$) coordinate (C);

where the ($...$) encloses the calc syntax, and the (<first node>)!<distance>!(<second node>) specifies the coordinate that lies at <distance> along the line from <first node> to <second node>.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,intersections}
\begin{document}
\begin{tikzpicture}
\fill (0,0) circle [radius=2pt] node (A) [label=A] {};
\fill (2,1) circle [radius=2pt] node (B) [label=B] {};
\fill (3.2,1) circle [radius=2pt] node (C) [label=C] {};
\fill (4,0.5) circle [radius=2pt] node (D) [label=D] {};
\draw [name path=AB] (A) -- ($(B)!-1cm!(A)$);
\draw [name path=CD] (D) -- ($(C)!-1cm!(D)$);

\fill [red,name intersections={of={AB and CD}}] (intersection-1) circle [radius=2pt];
\end{tikzpicture}
\end{document}

Answer 2

If you use tkz-euclide (documentation only in French, though), this is much easier:

\documentclass{article}

\usepackage{tkz-euclide}

\begin{document}
\begin{tikzpicture}

\tkzDefPoint(0,0){A} \tkzDefPoint(2,1){B}
\tkzDefPoint(3.2,1){C} \tkzDefPoint(4,0.5){D}

\tkzInterLL(A,B)(C,D) \tkzGetPoint{I}
\tkzDrawLines(A,I D,I)
\tkzDrawPoints[color=blue](A,B,C,D) \tkzDrawPoint[color=red](I)
\tkzLabelPoints[above](I,A,B,C,D)

\end{tikzpicture}
\end{document}

You could also write a piece of code to check the coordinates and automate the choice points for the \tkzDrawLines command and then use arbitrary coordinates, if you need, but I was too lazy to do it. :)

Answer 3

Consider this a side-answer to the original question, since the solution is provided using pstricks.

pstricks-add provides this point of intersection by means of the macro \psIntersectionPoint(<P0>)(<P1>)(<P2>)(<P3>){<node name>} where P0 and P1 define one straight line (say 0-1) and P2 and P3 define another straight line (say 2-3). <node name> is the name of the saved node at the intersection of 0-1 and 2-3. The following example is taken directly from the pstricks-add package documentation:

\documentclass{article}
\usepackage{pstricks-add}% http://ctan.org/pkg/pstricks-add}
\begin{document}
\psset{unit=0.5cm}
\begin{pspicture}(-5,-4)(5,5)
  \psaxes[labelFontSize=\scriptstyle,
    dx=2,Dx=2,dy=2,Dy=2]{->}(0,0)(-5,-4)(5,5)
  \psline[linecolor=red,linewidth=2pt](-5,-1)(5,5)
  \psline[linecolor=blue,linewidth=2pt](-5,3)(5,-4)
  \qdisk(-5,-1){2pt}\uput[-90](-5,-1){A}
  \qdisk(5,5){2pt}\uput[-90](5,5){B}
  \qdisk(-5,3){2pt}\uput[-90](-5,3){C}
  \qdisk(5,-4){2pt}\uput[-90](5,-4){D}
  \psIntersectionPoint(-5,-1)(5,5)(-5,3)(5,-4){IP}
  \qdisk(IP){3pt}\uput{0.3}[90](IP){IP}
  \psline[linestyle=dashed](IP|0,0)(IP)(0,0|IP)
\end{pspicture}
\end{document}