24

I am using TikZ for the first time, and I am experiencing some difficulty getting what exactly I want it to display. I am trying to draw the following picture:

wanted

Here is the code (I know it is not that good, but note it's my first time using TikZ):

\documentclass[paper=a4, fontsize=11pt]{scrartcl} % A4 paper and 11pt font size

\usepackage[T1]{fontenc}
\usepackage[english]{babel} % English language/hyphenation
\usepackage{amsmath,amsfonts,amsthm} % Math packages
\usepackage{graphicx}
\usepackage{mdframed}
\usepackage[ampersand]{easylist}
\usepackage{enumitem}
\usepackage{tikz}

\begin{document}
\begin{center}\begin{tikzpicture}[auto]
\draw[thick, ->] (-6,0) -- (10,0) node[anchor = north west] {x};
\draw[thick, ->] (0,-3) -- (0,11) node[anchor = south east] {y};
\node (P) at (-4,-2) {$P(-4;-2)$};
\node (Q) at (8,3) {$Q(8;3)$};
\node (R) at (4,10) {$R(4;10)$};
\node (N) at (2,7) {N};
\draw (P) to (Q);
\draw (R) to (Q);
\draw (P) to (R);
\draw (N) to (Q);
\end{tikzpicture}\end{center}
\end{document}

and it is currently giving me the following:

output

How can I

  • get the point names to not cut through the lines,
  • draw the line connecting QN to be indicated as a perpendicular line?
22

I'll just show one out of many possibilities (and yet another alternative for 90-degrees angle mark - for other angles have a look at the angles library in the manual for v3.00-)

\documentclass[tikz]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[scale=5]% Scale it rather using too big dimensions in centimeters
\coordinate[label=below:{$P(-4,-2)$}](p) at (-4mm,-2mm);
\coordinate[label=right:{$Q(8,3)$}](q) at (8mm,3mm);
\coordinate[label=above:{$R(4,10)$}] (r) at (4mm,10mm);

\draw (p) -- (q) -- (r) -- cycle; %Look at the tip of the triangle with cycle or (p)
% Here is some black magic; start from q and draw to a point
% which is at the place along the line from p to r but at the 
% place where q is projected on that line.
\draw (q) -- ($(p)!(q)!(r)$) coordinate (s);
\draw ($(s)!0.5mm!(q)$) coordinate (t) -- ($(t)!0.5mm!90:(q)$) --($(s)!0.5mm!(r)$);
\end{tikzpicture}
\end{document}

output

While you are developing your TikZ-fu, keep an eye on the package from our own Alain Matthes, called tkz-euclide. It makes these kind of drawings really really easy and structured. The only downside is that the manual is French to me (literally). But it is pretty self-explanatory nevertheless.

  • Thank you! :). Can you please show me how to get the "perpendicular block" at N from line QN as well? :) – user860374 Feb 15 '15 at 14:12
  • 1
    +1; I have just seen that page 146 of the TikZ manual v3.0 has the exact same solution. :) – Pier Paolo Feb 15 '15 at 14:22
  • This code is very easy to follow :). Thank you! :). It worked wonderfully! How can I label the "projected" line's point as N? – user860374 Feb 15 '15 at 14:32
  • 1
    @PierPaolo Apparently, I subconsciously memorized the TikZ manual. – percusse Feb 15 '15 at 14:52
  • 2
    @Dillon \draw (q) -- ($(p)!(q)!(r)$)node[draw,pos=1,sloped,anchor=south west]{}node[pos=1.05]{$N$};. – user11232 Feb 15 '15 at 15:12
17

With tkz-euclide:

\documentclass[11pt,a4paper]{article}
\usepackage{tkz-euclide}
\usetkzobj{all}
\begin{document}
\begin{tikzpicture}
\tkzInit[xmin=-5,xmax=9,ymin=-3,ymax=11]
\tkzAxeXY
%\tkzGrid
\tkzDefPoint[label=below:{$P(-4,-2)$}](-4,-2){P}
\tkzDefPoint[label=right:{$Q(8,3)$}](8,3){Q}
\tkzDefPoint[label=above:{$R(4,10)$}](4,10){R}
\tkzDrawSegments(P,Q Q,R R,P)
\tkzDefPointBy[projection=onto P--R](Q)
\tkzGetPoint{N}
\tkzLabelPoints[above left](N)
\tkzDrawPoints[color=red](P,Q,R,N)
\tkzDrawSegment(Q,N)
\tkzMarkRightAngle[fill=lightgray](Q,N,R)
\tkzLabelAngle[pos=1.3](Q,P,R){$\beta$}
\tkzMarkAngle[arc=l,size=1cm](Q,P,R)
\end{tikzpicture}
\end{document}

enter image description here

  • Thank you! :) . How can I label the point of intersection of the projected line onto PR as N? – user860374 Feb 15 '15 at 14:35
  • 1
    @Dillon Please see the edit. – user11232 Feb 15 '15 at 14:44
8

And for comparison, with Metapost.

enter image description here

Note that unlike the warning at the top of the OP diagram, this one is drawn to scale...

prologues := 3;
outputtemplate := "%j%c.eps";

beginfig(1);
u := 5mm;

% axes
path xx, yy;
xx = (5 left -- 9 right) scaled u;
yy = (3 down -- 11 up) scaled u;
drawarrow xx withcolor .5 white; 
drawarrow yy withcolor .5 white;
label.rt (btex $x$ etex, point 1 of xx);
label.top(btex $y$ etex, point 1 of yy);

% define the points
pair M, N, P, Q, R;
P = (-4, -2) scaled u;
Q = ( 8,  3) scaled u;
R = ( 4, 10) scaled u;
M = yy intersectionpoint (P--R);
N = whatever[P,R]; (N-Q) dotprod (R-P) = 0;
%
% mark the right angle
draw unitsquare scaled 5 rotated angle (Q-N) shifted N withcolor .5 white;
% draw the lines
draw P--Q--R--cycle; draw Q--N;
% add the labels
label.ulft(btex $M$          etex, M);
label.ulft(btex $N$          etex, N);
label.top (btex $R\,(4;10)$  etex, R);
label.rt  (btex $Q\,(8;3)$   etex, Q);
label.lft (btex $P\,(-4;-2)$ etex, P);
% label the angle along the bisector
label(btex $\beta$ etex, P + 20 unitvector(Q+R-2P));
endfig;
end.

A little thought about the geometry here, shows us that since M is on the y-axis it is half way between P and R in the x-direction, so it must be the midpoint of P--R. It therefore has co-ordinates (0,4); and the distance from M to Q is therefore sqrt(8^2+1^2)=sqrt(65), but this is the same as the distance from R to Q, which is sqrt(4^2+7^2)=sqrt(65); hence QNR and QNM are congruent triangles, and therefore N is the midpoint of R--M with coordinates (2,7). You can use Metapost equation system to confirm this; if you add

M = (0,4) scaled u; N = (2,7) scaled u;

after the implicit definitions already given, then MP gives no error.

  • 1
    how do you know if it's to scale when we do not know the units of the original? ;-) – Paul Gessler Feb 15 '15 at 19:25
8

A PSTricks solution using the pst-eucl package:

\documentclass{article}

\usepackage{pst-eucl,pstricks-add}
\psset{unit = 0.76, PointSymbol = none}

\begin{document}

\begin{pspicture}(-5.3,-3.2)(10,11)
  \pnodes(-4,-2){P}(8,3){Q}(4,10){R}(0,0){O}(0,10){Y}
  \psaxes{->}(0,0)(-5.3,-3.2)(9.5,10.5)[$x$,0][$y$,90]
  \pspolygon(P)(Q)(R)
  \uput[270](P){$P(-4,-2)$}
  \uput[350](Q){$Q(8,3)$}
  \uput[90](R){$R(4,10)$}
  \pstInterLL{P}{R}{O}{Y}{M}
  \pstProjection{P}{R}{Q}[N]
  \pstLineAB{Q}{N}
  \pstLineAB{P}{N}
  \pstRightAngle{P}{N}{Q}
  \pstMarkAngle{Q}{P}{R}{$\beta$}
\end{pspicture}

\end{document}

output

4

With the unfairly unkwnown mfpic package. It is a wide set of (La)TeX macros providing a very convenient interface to either METAFONT or MetaPost, in this case the latter.

Since most are not familiar with this package, I've put a fair number of comments in the following code, hence its relative length.

For those who know a bit about MetaPost, mfpic can also embed raw MetaPost instructions which are sometimes more convenient. As I did here to locate the points M and N (which is done by MetaPost in its trademark implicit way), and to draw the right angle mark upon N (thanks to MetaPost transformers).

\documentclass{scrartcl}
% MetaPost instead of Metafont as drawing program and labels manager.
% Bounding box based on actual picture dimensions, not on the axes dimensions.
\usepackage[metapost, mplabels, truebbox]{mfpic}
% LaTeX preamble given to MetaPost for its labels management
% (corresponds to the verbatimtex ... etex flags of MetaPost)
\mfpverbtex{%&latex
    \documentclass{scrartcl}
    \begin{document}}
\setlength{\mfpicunit}{.5cm}
\opengraphsfile{\jobname}
\begin{document}
\begin{mfpic}[1]{-5}{9.5}{-3}{11}
    % Points definitions. For MetaPost they are local pairs.
    \setmfpair{P}{(-4, -2)}
    \setmfpair{Q}{(8, 3)}
    \setmfpair{R}{(4, 10)}
    % Point M computed by MetaPost as intersection of y-axis and straight line (PQ)
    \setmfpair{M}{(P -- R) intersectionpoint (origin -- (0, \ymax))}
    % Point N computed by MetaPost as intersection of line (PQ)
    % and the straight line perpendicular to (PQ) going through Q
    \mfsrc{save N; pair N; N = whatever[P, R] = whatever[Q, Q + (R-P) rotated 90];}
    % Mark angle beta with the convenient \arc macro of mfpic
    \store{mark_angle}\arc[a]{P, 0.8, angle(Q-P), angle(R-P)}
    % Mark right angle on N with help of MetaPost transformers
    \setmfvariable{path}{mark_right_angle}
        {((1, 0) -- (1, 1) -- (0, 1)) zscaled 0.3unitvector(Q-N) shifted N}
    % Actual drawings
    \polygon{P, Q, R}
    \lines{Q, N}
    \doaxes{xy}
    \mfobj{mark_angle}
    \mfobj{mark_right_angle}
    % Labels
    \tlpointsep{2bp} % Offset
    \tlabels{[tc]{(\xmax, 0)}{$x$} [cr]{(0, \ymax)}{$y$}
        [cr]{P}{$P(-4, -2)$} [cl]{Q}{$Q(8, 3)$} [bc]{R}{$R(4, 10)$}
        [br]{M}{$M$} [cr]{N}{$N$} [bl]{point 0.4 of mark_angle}{$\beta$}}
\end{mfpic}
\closegraphsfile
\end{document}

The .tex file is to be typeset by (pdf)LaTeX first, then the resulting .mp file by MetaPost, and finally the LaTeX file by (pdf)LaTeX again, to produce the figure below.

enter image description here

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