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繪製垂直於給定直線的特定長度的線段

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繪製垂直於給定直線的特定長度的線段

我有一個典型的幾何結構，我想用它來繪製TikZ。下面的程式碼給了一個線段PQ。如何在垂直並穿過該線段下方單位的線上TikZ繪製一個點？我認為我必須在序言中做到這一點。 （我想將這個點標記為R並繪製直角三角形。）如果我想用它來指示該三角形是直角三角形，我必須在序言中添加嗎？PQP2\sqrt{2}PQ\usetikzlibrary{calc}QPR\tkzMarkRightAngle(Q,P,R);\usepackage{tkz-euclide}

\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}

答案1

檢查程式碼中的解釋性註解：

\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}

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答案2

為了回答您的問題，tikz 通過 pgfmathparse 運行幾乎所有內容，它看起來更像 C 代碼：2*sqrt(5)，順便說一句，默認比例超過 4 厘米。 ($(P)!1!90:(Q)$) 表示從 (P) 開始，從 (Q) 方向走 1 倍距離到 (Q) 90 度（逆時針）。

\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}

計算用量

答案3

事實上，你並不是被迫使用\usepackage{tkz-euclide}.它只為您提供了\tkzMarkRightAngle(Q,P,R)比 pure 更簡單的構造tikz，但這也可以相對輕鬆地實現，僅使用tikz@HarishKumar 所做的那樣。此外，使用styles 表示軸和點可以使程式碼更加清晰。

\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}

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答案4

對於那些有興趣的人，一種（除其他外）這樣做的方法元郵報。

使 PR 垂直於 PQ 且位於 PQ 下方距 P 個單位的點 R2\sqrt{2}由下列指令給出（u此處為單位長度cm）：

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

anglebetweenMetaPost 的宏梅塔芬如果參數設為 2，第三個參數是標籤，此處為空字串）。P--Q（格式允許在兩個相交線段之間輕鬆繪製直角標記（預設為 20 pt）。P--Ranglemethodanglelength

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

看Metafun 手冊，p。 279，詳細介紹。

\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}

用LuaLaTeX排版。輸出：

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