How to plot polygon using TikZ

Question

I would like to plot a 2D $x_1 - x_2$ coordinate system and a shaded polygon specified by

$ −3x_1 + 4x_2 \leq 4, 3x_1 + 2x_2 \leq 11, 2x_1 − x_2 \leq 5, x_1 , x_2 \geq 0 $

How to draw this using TikZ?

Thanks and regards!

---

Update:

I just learned something from this post (or if you have better idea, please don't hesitate to reply).

Here is my plot and I was wondering how to add different shades to the polygon enclosed by the solid lines and to the polygon enclosed by the dashed lines?

Here is my code:

\documentclass{article}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}
    \draw[very thin,color=gray] (-0.1,-0.1) grid (4.9,3.1);    
    \draw[->] (-0.2,0) -- (4.2,0) node[right] {$x_1$};
    \draw[->] (0,-0.2) -- (0,3.2) node[above] {$x_2$};

    \draw[domain=0:3] plot (\x,{1+ 0.75 * \x}) node[above right] {$-3x_1+4x_2 =4$};
    \draw[domain=1.5:4] plot (\x,{5.5 - 1.5 * \x}) node[below right] {$3x_1 + 2x_2 = 11$};
    \draw[domain=2:4] plot (\x,{-5+2 * \x}) node[below right] {$2x_1 - x_2 =5$}; 

    \node at (2,3) {(2, 2.5)};
    \node at (3.5,1) {(3, 1)};
    \node at (3,-0.25) {(2.5, 0)};    
    \node at (-0.25,-0.25) {(0, 0)};
    \node at (-0.5,1) {(0, 1)};

    \draw[dashed] (0,1) -- (2,2);
    \draw[dashed] (2,2) -- (3,1);
    \draw[dashed] (3,1) -- (2,0); 
    \draw[dashed] (2,0) -- (0,0); 
    \draw[dashed] (0,0) -- (0,1); 

    \node at (2,1.5) {(2, 2)};
    \node at (2,-0.25) {(2, 0)};

\end{tikzpicture}
\end{document} 

Answer 1

Here's another way to do it without using intersections but using path clipping. Each path we draw, we also define a clip against this path. This does mean "drawing" the path twice: once to draw and once to clip against (and I don't see a quick way of merging these two since they have to happen at different times). One thing I like about this approach is that what the graphical package is doing is precisely what the mathematics is doing. That is, each inequality specifies a "clipping" of the plane, saying "After this, we're only interested in one side of this line and we throw away everything else.". That's exactly what a clip does.

\documentclass{article}
\pagestyle{empty}
\usepackage[svgnames]{xcolor}
\usepackage{tikz}
\usepackage{mathtools}

\def\nudge{.5}

\tikzset{axis/.style={ultra thick, Red!75!black, -latex, shorten <=-\nudge cm, shorten >=-2*\nudge cm}}
\tikzset{line/.style={thick,Green}}

\begin{document}
\begin{tikzpicture}
\draw[axis] (0,0) -- (4,0) node[right=2* \nudge cm] {\(x_1\)};
\draw[axis] (0,0) -- (0,4) node[above=2*\nudge cm] {\(x_2\)};
\begin{scope}
\clip (-\nudge ,-\nudge) rectangle (4+\nudge,4+\nudge);
\draw[line] (0,1) -- (4,4) coordinate (ineq1);
\draw[line] (0,5.5) -- (4,-.5) coordinate (ineq2);
\draw[line] (0,-5) -- (4,3) coordinate (ineq3);
\begin{scope}
\clip (0,1) -- (4,4) |- (0,0);
\clip (0,5.5) -- (4,-.5) -| (0,0);
\clip (0,-5) -- (4,3) |- (4,4) -| (0,0);
\fill[Red,opacity=.5] (0,0) rectangle (4,4);
\end{scope}
\draw[dashed,line] (0,1) -- (2,2) -- (3,1) -- (2,0);
\clip (0,1) -- (2,2) -- (3,1) -- (2,0) -| (0,1);
\fill[Blue,opacity=.5] (0,0) rectangle (4,4);
\end{scope}
\node[above right] at (ineq1) {\(\mathllap{-}3 x_1 + 4 x_2 = 4\)};
\node[below right] at (ineq2) {\(3 x_1 + 2 x_2 = 11\)};
\node[above right] at (ineq3) {\(2 x_1 - x_2 = 5\)};
\foreach \coord/\adj in {
  {(2,2)}/right,
  {(0,1)}/left,
  {(0,0)}/below left,
  {(2,0)}/below left,
  {(2.5,0)}/below right,
  {(3,1)}/right,
  {(2,2.5)}/right%
} {
  \fill \coord circle (2pt) node[\adj] {\coord};
}
\end{tikzpicture}
\end{document}

Other aspects of this answer:

1. We use clipping to avoid having to work out the formulae for the lines too carefully (and we draw these as simple lines rather than using the plot function).
2. We use node-coordinates when we want to label the lines since putting the labels directly on the lines wouldn't work with the clipping.
3. Note the use of \mathllap to adjust the placement of one of the labels (that's why mathtools is included).
4. Note the use of a negative shortening of the axes.

Answer 2

Here is one way to do it. The code uses intersections to calculate the vertices of the polygon (note that the intersection library is only available starting from TikZ v2.10). The labels are generated using the let operation, which gives access to coordinates of nodes (thank you Antal). Filling paths is described in section 15.4 “Filling a Path” of the TikZ v2.10 manual. You could also apply various patterns.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections,positioning,calc}
\begin{document}

% print a point given by two coordinates in pt (output is in cm)
\newcommand*\printpoint[2]{(%
    \pgfmathparse{0.03514598035*#1}\pgfmathprintnumber{\pgfmathresult}, %
    \pgfmathparse{0.03514598035*#2}\pgfmathprintnumber{\pgfmathresult})%
}

\begin{tikzpicture}
    % grid and axes
    \draw[very thin,color=gray] (-0.1,-0.1) grid (4.1,3.1);    
    \draw[->,name path=xaxis] (-0.2,0) -- (4.2,0) node[right] {$x_1$};
    \draw[->,name path=yaxis] (0,-0.2) -- (0,3.2) node[above] {$x_2$};

    % lines  
    \draw[name path=line1,domain=0:3] plot (\x,{1+ 0.75 * \x}) node[above right] {$-3x_1+4x_2 =4$};
    \draw[name path=line2,domain=1.5:4] plot (\x,{5.5 - 1.5 * \x}) node[below right] {$3x_1 + 2x_2 = 11$};
    \draw[name path=line3,domain=2:4] plot (\x,{-5+2 * \x}) node[below right] {$2x_1 - x_2 =5$}; 

    % calculate intersection points
    \node[name intersections={of=line1 and line2}] (a) at (intersection-1) {};
    \node[name intersections={of=line2 and line3}] (b) at (intersection-1) {};
    \node[name intersections={of=line3 and xaxis}] (c) at (intersection-1) {};
    \node (d) at (0,0) {};
    \node[name intersections={of=yaxis and line1}] (e) at (intersection-1) {};

    % draw the big polygon    
    \filldraw[ultra thick,fill=green!80!black,fill opacity=0.4] (a.center) -- (b.center) -- (c.center) -- (d.center)  -- (e.center) -- cycle;

    % label the vertices
    \path let \p0 = (a) in node [left=0.1cm of a] {\printpoint{\x0}{\y0}};
    \path let \p0 = (b) in node [right=0cm of b] {\printpoint{\x0}{\y0}};
    \path let \p0 = (c) in node [below right=0cm and -0.1cm of c.center] {\printpoint{\x0}{\y0}};
    \path let \p0 = (d) in node [below left=0cm of d.center] {\printpoint{\x0}{\y0}};
    \path let \p0 = (e) in node [left=0cm of e.center] {\printpoint{\x0}{\y0}};

    % draw the small polygon   
    \filldraw[thick,dashed,fill=blue,fill opacity=0.4] (0,1) -- (2,2) -- (3,1) -- (2,0) -- (0,0) -- cycle; 
\end{tikzpicture}
\end{document}

Answer 3

My solution uses my new package tkz-fct and tkz-euclide but it's not necessary to use tkz-fct because it's possible with straight lines to get points without gnuplot :( . If you want directly the lines you defin a point with tkzDefpoint(x,y){Name} and the you can draw a line with tkzDrawLine(A,B) . Here my code with tkz-fct

remark : for a line samples =2 is enough and it's possible !

 \documentclass{article}
    \usepackage{tkz-euclide,tkz-fct}
    \usetkzobj{all}

    \begin{document} 
    \begin{tikzpicture}
         \tkzInit \tkzClip[space=.5]
       \tkzDefPoint(0,0){O} 
       \tkzDrawX[label=$x_1$] \tkzDrawY[label=$x_2$]
       \tkzFct[color   = red,domain =0:10,samples=2]{4+0.75*x}
       \tkzDefPointByFct(0) \tkzGetPoint{A1} 
       \tkzDefPointByFct(10) \tkzGetPoint{B1} 
       \tkzFct[color   = blue,domain =0:10,samples=2]{11-1.5*x}
       \tkzDefPointByFct(0) \tkzGetPoint{A2} 
       \tkzDefPointByFct(10) \tkzGetPoint{B2}  
       \tkzFct[color   = green,domain =0:10,samples=2]{2*x-5}
       \tkzDefPointByFct(2.5) \tkzGetPoint{A3} 
       \tkzDefPointByFct(10) \tkzGetPoint{B3} 
       \tkzInterLL(A1,B1)(A2,B2) \tkzGetPoint{I12} 
       \tkzInterLL(A1,B1)(A3,B3) \tkzGetPoint{I13}
       \tkzInterLL(A3,B3)(A2,B2) \tkzGetPoint{I23}
       \tkzFillPolygon[color=magenta!30](O,A3,I23,I12,A1)
       \tkzDrawPolygon[color=magenta](O,A3,I23,I12,A1) 
       \tkzDrawPoints(A1,A2,A3,I12,I13,I23)
        \end{tikzpicture} 
\end{document} 

Alain Matthes