Drawing a line segment of a specific length that is perpendicular to a given line

I have a typical geometric construction that I want to draw using TikZ. The following code gives a line segment PQ. How do I get TikZ to draw a point on the line perpendicular to PQ and through P that is 2\sqrt{2} units below the line segment PQ? I think that I have to have \usetikzlibrary{calc} in the preamble to do this. (I want to label this point R and draw right triangle QPR.) If I want to use \tkzMarkRightAngle(Q,P,R); to indicate that the triangle is a right triangle, do I have to have \usepackage{tkz-euclide} in the preamble?

\documentclass{amsart}
\usepackage{tikz}
\usetikzlibrary{shapes,positioning,intersections,quotes}

\begin{document}

\begin{tikzpicture}

\draw[yellow, line width=0.1pt] (-1.75,-1.75) grid[xstep=0.5, ystep=0.5]  (2.75,1.75);
\draw[draw=gray!30,latex-latex] (0,1.75) +(0,0.25cm) node[above right] {$y$} -- (0,-1.75) -- +(0,-0.25cm);
\draw[draw=gray!30,latex-latex] (-1.75,0) +(-0.25cm,0) -- (2.75,0) -- +(0.25cm,0) node[below right] {$x$};

\node[outer sep=0pt,circle, fill,inner sep=1.5pt,label={[fill=white]left:$P$}] (P) at (-1,-1) {};
\node[outer sep=0pt,circle, fill,inner sep=1.5pt, label={[fill=white]right:$Q$}] (Q) at (2,1) {};

\draw[green!20!white] (P) -- (Q);

\end{tikzpicture}
\end{document}
• – Claudio Fiandrino Apr 20 '15 at 14:56
• Check page 145 of the manual (calc ltikzlibrary). – John Kormylo Apr 20 '15 at 16:01
• @John Kormylo I made the grid more expansive and included the code \coordinate (R) at ($(P)!2\sqrt{5}!90:(Q)$); and got an error message. It seems to me that TikZ could not interpret \sqrt{5} as part of a measurement. I changed the command to \coordinate (R) at ($(P)!1!90:(Q)$); and nothing was plotted. I think that I need to have cm as the units. Isn't cm the default unit assigned to a number? I changed the command to \coordinate (R) at ($(P)!1cm!90:(Q)$); and nothing was plotted. – user74973 Apr 21 '15 at 0:56
• @John Kormylo I would like to have a dot put at R. Why isn't \node[outer sep=0pt,circle, fill,inner sep=1.5pt,label={[fill=white]above left:$R$}] (R) at R {}; putting a dot at R? – user74973 Apr 21 '15 at 0:57
• @John Kormylo By the way, how do I know whether -90 or 90 will make the coordinate R below line segment PQ? – user74973 Apr 21 '15 at 0:58

Check the code for explanatory comments:

\documentclass{amsart}
\usepackage{tikz}

%% you need the following 2 lines to use \tkzMarkRightAngle
%\usepackage{tkz-euclide}
%\usetkzobj{all}

\usetikzlibrary{shapes,positioning,intersections,quotes,calc}

\begin{document}

\begin{tikzpicture}

\draw[yellow, line width=0.1pt] (-1.75,-5) grid[xstep=0.5, ystep=0.5]  (2.75,1.75);
\draw[draw=gray!30,latex-latex] (0,1.75) +(0,0.25cm) node[above right] {$y$} -- (0,-1.75) -- +(0,-0.25cm);
\draw[draw=gray!30,latex-latex] (-1.75,0) +(-0.25cm,0) -- (2.75,0) -- +(0.25cm,0) node[below right] {$x$};

\node[outer sep=0pt,circle, fill,inner sep=1.5pt,label={[fill=white]left:$P$}] (P) at (-1,-1) {};
\node[outer sep=0pt,circle, fill,inner sep=1.5pt, label={[fill=white]right:$Q$}] (Q) at (2,1) {};

\draw[green!20!white] (P) -- (Q);

%% the perpendicular
\pgfmathparse{2*sqrt(5)}
\node[outer sep=0pt,circle, fill,inner sep=1.5pt,label={[fill=white]left:$R$}] (R) at ($(P)!\pgfmathresult cm! -90:(Q)$) {};
\draw[green!20!white] (P) -- (R) -- (Q);

%% right angle with tkz-euclide
%\coordinate (p) at (P);
%\tkzMarkRightAngle[color=green!20!white](Q,p,R)
%\fill (p) circle (2.1pt);  %% to make the dot above right angle again.

\coordinate (a) at ($(P)!5mm! -45:(Q)$);
\draw[green!20!white] (a) -- ($(P)!(a)!(Q)$);
\draw[green!20!white] (a) -- ($(P)!(a)!(R)$);

\end{tikzpicture}
\end{document} • I understand that the command \coordinate (a) at ($(P)!5mm! -45:(Q)$); is instructing TikZ to put a coordinate on the graph and to call it "a." With -45, isn't the command instructing TikZ to put the point at a distance of 5mm from P and at an angle of 45 degrees in the clockwise direction from line segment PQ? – user74973 Apr 21 '15 at 18:21
• @user74973 Yes. – user11232 Apr 21 '15 at 23:20

To answer your questions, tikz runs almost everything through pgfmathparse which looks more like C code: 2*sqrt(5) which BTW is over 4cm at the default scale. ($(P)!1!90:(Q)$) means start at (P) and go 1 times the distance to (Q) 90 degrees (counter-clockwise) from the direction to (Q).

\documentclass{amsart}
\usepackage{tikz}
\usetikzlibrary{shapes,positioning,intersections,quotes,calc}

\begin{document}

\begin{tikzpicture}

\draw[yellow, line width=0.1pt] (-1.75,-1.75) grid[xstep=0.5, ystep=0.5]  (2.75,1.75);
\draw[draw=gray!30,latex-latex] (0,1.75) +(0,0.25cm) node[above right] {$y$} -- (0,-1.75) -- +(0,-0.25cm);
\draw[draw=gray!30,latex-latex] (-1.75,0) +(-0.25cm,0) -- (2.75,0) -- +(0.25cm,0) node[below right] {$x$};

\node[outer sep=0pt,circle, fill,inner sep=1.5pt,label={[fill=white]left:$P$}] (P) at (-1,-1) {};
\node[outer sep=0pt,circle, fill,inner sep=1.5pt, label={[fill=white]right:$Q$}] (Q) at (2,1) {};

\draw[green!20!white] (P) -- (Q);

\coordinate (R) at ($(P)!2cm*sqrt(5)!-90:(Q)$);
\node[outer sep=0pt,circle, fill,inner sep=1.5pt, label={[fill=white]right:$R$}] at(R) {};

\end{tikzpicture}
\end{document} • You said that "($(P)!1!90:(Q)$) means start at (P) and go 1 times the distance to (Q) 90 degrees (counter-clockwise) from the direction to (Q)." This code will give me a right triangle. Right? It won't be the one intended - it will draw the height from the base PQ at Q. – user74973 Apr 21 '15 at 16:35

Actually, you are not forced to use \usepackage{tkz-euclide}. It only gives you an easier \tkzMarkRightAngle(Q,P,R) construct than pure tikz, but this can also be achieved relatively easily using only tikz as @HarishKumar did it. Also, using styles for axes and points makes the code cleaner.

\documentclass[tikz]{standalone}
\usetikzlibrary{calc}
\begin{document}\small

\begin{tikzpicture}[outer sep=0pt,p/.style={circle, fill,inner sep=1.5pt},ax/.style={draw=gray!50,latex-latex}]
\draw[yellow,line width=0.1pt] (-2.75cm,-1.75cm) grid[xstep=0.5, ystep=0.5]  (2.75cm,1.75cm);
\draw[ax](0,1.75cm) +(0,0.25cm) node[above] {$y$} -- (0,-1.75cm) -- +(0,-0.25cm);
\draw[ax](-2.75cm,0) +(-0.25cm,0) -- (2.75cm,0) -- +(0.25cm,0) node[right] {$x$};
\coordinate[p,label={[fill=white]below:$P$}] (P) at (-1cm,-1cm);
\coordinate[p,label={[fill=white]right:$Q$}] (Q) at (2cm,1cm);
\coordinate[p,label={[fill=white]left :$R$}] (R) at ($(P)!{2cm*sqrt(2)}!90:(Q)$);
\draw (P)--(R)--(Q)--(P)--cycle;
\coordinate (a) at ($(P)!4mm!45:(Q)$);
\draw ($(P)!(a)!(Q)$) -- (a) -- ($(P)!(a)!(R)$);
\end{tikzpicture}

\end{document} • I looked on page 216 of the manual at http://texdoc.net/texmf-dist/doc/generic/pgf/pgfmanual.pdf and it only says that specifying coordinates follows at. Now, you tell me that you can locate a point with $(P)!{2cm*sqrt(2)}!90:(Q)$. That will be helpful in coding. Thanks. – user74973 Apr 21 '15 at 16:22
• I see how you got a length of 2\sqrt{2} - you scaled 2cm by \sqrt{2}. Isn't the default unit centimeters? Why can't you get the same picture with $(P)!{2*sqrt(2)}!90:(Q)$? – user74973 Apr 21 '15 at 16:28

For those who are interested, a way (among others) of doing it with MetaPost.

The point R such that PR is perpendicular to PQ and located 2\sqrt{2} units away from P below PQ, is given by the following instruction (u being the unit length, here cm):

R = P + 2u*sqrt2*unitvector(Q-P) rotated -90;

The anglebetween macro of MetaPost's Metafun format allows to draw right angle marks quite easily between the two intersecting segments P--Q and P--R (the third argument being the label, here an empty string), if the anglemethod parameter is set to 2. With the default value of this parameter, 1, the mark would be a circular arc. The desired length of the mark should be given as argument to another parameter, anglelength (which defaults to 20 pt).

anglemethod := 2; anglelength := 2mm;
draw anglebetween(P--Q, P--R, "");

See the Metafun manual, p. 279, for a detailed presentation.

\documentclass[border=2mm]{standalone}
\usepackage{luamplib}
\mplibsetformat{metafun}
\mplibtextextlabel{enable}
\begin{document}
\begin{mplibcode}
% Axes parameters
u := cm; % Unit length
xmin := -1.75u; xstep := .5u; xmax := 2.75u;
ymin := -5u; ystep := xstep; ymax := 1.75u;
% Triangle summits
pair P, Q, R; P = u*(-1, -1); Q = u*(2, 1);
R = P + 2u*sqrt2*unitvector(Q-P) rotated -90;
beginfig(1);
% Grid
drawoptions(withcolor yellow);
for i = ceiling(xmin/xstep) upto floor(xmax/xstep):
draw (i*xstep, ymin) -- (i*xstep, ymax);
endfor
for j = ceiling(ymin/ystep) upto floor(ymax/ystep):
draw (xmin, j*ystep) -- (xmax, j*ystep);
endfor
% Axes
drawoptions(withcolor .8white);
drawarrow (xmin, 0) -- (xmax, 0);
drawarrow (0, ymin) -- (0, ymax);
% Triangle
drawoptions(withcolor green);
path triangle; triangle = P--Q--R--cycle; draw triangle;
% Right-angle mark of length 2 mm (and no label)
anglemethod := 2; anglelength := 2mm;
draw anglebetween(P--Q, P--R, "");
% Labels
drawoptions();
label.bot("$x$", (xmax, 0)); label.lft("$y$", (0, ymax));
dotlabel.lft("$P$", P); dotlabel.rt("$Q$", Q); dotlabel.bot("$R$", R);
endfig;
\end{mplibcode}
\end{document}

To be typeset with LuaLaTeX. Output: 