# How can I connect a random point in a smooth plot with the axis?

I have a random smooth drawing (just smoothly connecting dots) and want to highlight a point between the dots. Highlighting should happen by connecting the point on the graph with its coordinate on the x-axis.

Here is a MWE that I would like to keep as it is

\documentclass{standalone}

\usepackage{tikz}
\usepackage{pgfplots}

\begin{document}
\begin{tikzpicture}
\begin{axis}[ytick=\empty,xtick=\empty,
axis on top=false,xmin=0, xmax=10, ymin=0, ymax=5,
axis x line=left, axis y line=left, clip=false]
\node at (rel axis cs:1,0) [above, anchor=north west] {$x$};
\node at (rel axis cs:0,1) [above, anchor=west] {$y$};
(0,0)
(2,3)
(5,1.2)
(5.8,4)
(8,3.8)
(10,3.5)
} coordinate [pos=.9] (prior) ;

\end{axis}
\end{tikzpicture}
\end{document}


I want to connect the coordinate prior with the x-axis using a line parallel to the y-axis. My take was this

\draw[dotted] (prior)-- (rel axis cs:0.9,0);


However, for some reason unclear to me, this is not the right coordinate (it is a bit to the right). Now I wonder, how can I asses the right coordinate? The function plots a range from 0:10 so my math skills would hinge that .9 of this should also be .9 of the length of the x-axis. However, this seems to be incorrect. Has anybody an idea what the right coordinate is? Or even simpler: How can I connect the point (prior) with the x-axis in the shortest possible way?

I tried to work with the solution suggested in an answer to this question:Smooth pgfplots. The problem is that I do not know the y-coordinate of the point prior.

For completeness here is the output it produces (never mind the different labels of the axis)

So you are searching for this?

The reason why pos=0.9 is not at x=0.9 is, that indeed the coordinate is placed at 90% of the length of the line. And because there is a lot of "up and down" at the start of the line (i.e. at low x values), this part of the line is "longer" than the part at the end of the line (i.e. at high x values), which is almost a horizontal straight line.

(Just for completeness: So if your intention was to place a node at x=0.9 on the line than this is currently not possible directly using the pos feature, but there is already a feature request for that in the PGFPlots Tracker. But of course also this can be done indirectly e.g. using the intersections library from TikZ.)

% used PGFPlots v1.14
\documentclass[border=5pt]{standalone}
\usepackage{pgfplots}
\pgfplotsset{
% use this compat' level or higher so there is no need any more to prefix
% TikZ coordinates with axis cs:'
compat=1.11,
}
\begin{document}
\begin{tikzpicture}
\begin{axis}[
xtick=\empty,
ytick=\empty,
xmin=0,
xmax=10,
ymin=0,
ymax=5,
axis lines=center,
% (moved the axis labels here)
xlabel={$x$},
ylabel={$y$},
xlabel style={anchor=north west},
ylabel style={anchor=north east},
]

(0,0)
(2,3)
(5,1.2)
(5.8,4)
(8,3.8)
(10,3.5)
}
coordinate [pos=.9] (prior)
;

% to draw a vertical line from prior' down to the x axis use the |-' operator and state an arbitrary x value, but 0 as y value
\draw [dotted] (prior) -- (prior |- 0,0);
\end{axis}
\end{tikzpicture}
\end{document}


• Great, thanks! That was exactly what I've been looking for. And luckily, I do not care about the x coordinate so, the solution you proposed works perfect for me – johaschn Apr 22 '17 at 10:04
• Just out of curiosity, how does the |- operator work? I played around a bit and thought I could also draw a horizontal line (parallel to the axis) using \draw (prior) -- (0,prior |- 0) (or variants of this), but that seems not to work. – johaschn Apr 22 '17 at 18:20
• @johaschn, it works like "use the x coordinate of the point given at the | side and use the y coordinate of the point given at the - side. But 0 isn't a valid coordinate ... Use \draw (prior) -- (prior -| 0,0) to draw a horizontal line to the y axis. – Stefan Pinnow Apr 22 '17 at 18:33