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将整个 TikZ 图片垂直移动到对齐环境的一部分

tag-icon • tag-icon tikz-pgf
将整个 TikZ 图片垂直移动到对齐环境的一部分

我正在尝试将 TikZ 图片向上移动，以便消除解决方案占用的空间。我怎样才能将 TikZ 图片向上移动（一点点）进入对齐环境？仅供参考，以下是考试问题。

“从 $(-2,1)$ 到点 $P$ 的线段中点是 $(1, -1)$。确定 $P$ 的坐标。”

以下是我对考试问题的解答。

Let $P = (x,y)$. We need to find $x$ and $y$. We have
\begin{align*}
\text{Midpoint} = \biggl( \frac{-2 + x}{2} , \frac{1 + y}{2} \biggr)
\quad   &\implies \quad
(1,-1) = \biggl( \frac{-2 + x}{2} , \frac{1 + y}{2} \biggr)\\
&\implies \quad
\begin{dcases}
    1 = \frac{-2 + x}{2}\\
    -1 = \frac{1 + y}{2}
\end{dcases}\\
&\implies \quad
\begin{dcases}
    2 = -2 + x\\
    -2 = 1 + y
\end{dcases}
\quad \implies \quad
\begin{dcases}
    x = 4\\
    y = -3.
\end{dcases}
\end{align*}
%\begin{minipage}[t]{5cm}{        %DO I NEED THIS?
\begin{tikzpicture}[>=latex',scale=0.5]
% Draw grid lines
\draw[help lines] (-2.5,-3.5) grid (4.5,1.5);
% Draw x-axis
\draw[very thick,->] (-3,0) -- (5.5,0)
    node[right] {\large $x$}; 
% Draw y-axis
\draw[very thick, ->] (0,-4) -- (0,2.5) 
node[above] {\large $y$};
% Special points
\node[above, fill=SolutionColor] at (-2,1) {$(-2,1)$};
\fill (-2,1) circle (4pt);
\node[right, fill=SolutionColor] at (1.5,-1) {$(1,-1)$};
\fill (1,-1) circle (4pt);
\node[right, fill=SolutionColor] at (4,-3) {$P$};
\draw (-2,1) -- (1,-1) -- (4,-3);
\fill[red] (4,-3) circle (4pt);
\end{tikzpicture}
%\end{minipage}%    DO I NEED THIS?

在此处输入图片描述

答案1

这是 Martin Scharrer 的优秀软件包所要做的工作adjustbox：它允许（在许多其他事情中）修剪一个盒子的顶部（该盒子可以包含许多其他东西，其中包括tikzpicture），因此它的官方高度小于内容的实际高度。

如果你把你的包裹起来tikzpicture，\begin{adjustbox}{trim=0 0 0 2.5cm} ... \end{adjustbox}你会得到

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{arrows}
\usepackage{amsmath}
\usepackage{mathtools}
\usepackage{adjustbox}

\begin{document}
Let $P = (x,y)$. We need to find $x$ and $y$. We have
\begin{align*}
\text{Midpoint} = \biggl( \frac{-2 + x}{2} , \frac{1 + y}{2} \biggr)
\quad   &\implies \quad
(1,-1) = \biggl( \frac{-2 + x}{2} , \frac{1 + y}{2} \biggr)\\
        &\implies \quad
\begin{dcases}
    1 = \frac{-2 + x}{2}\\
    -1 = \frac{1 + y}{2}
\end{dcases}\\
        &\implies \quad
\begin{dcases}
    2 = -2 + x\\
    -2 = 1 + y
\end{dcases}
\quad \implies \quad
\begin{dcases}
    x = 4\\
    y = -3.
\end{dcases}
\end{align*}

\begin{adjustbox}{trim=0 0 0 2.5cm}%
\begin{tikzpicture}[>=latex',scale=0.5]
% Draw grid lines
\draw[help lines] (-2.5,-3.5) grid (4.5,1.5);
% Draw x-axis
\draw[very thick,->] (-3,0) -- (5.5,0)
    node[right] {\large $x$}; 
% Draw y-axis
\draw[very thick, ->] (0,-4) -- (0,2.5) 
node[above] {\large $y$};
% Special points
\node[above, fill=white] at (-2,1) {$(-2,1)$};
\fill (-2,1) circle (4pt);
\node[right, fill=white] at (1.5,-1) {$(1,-1)$};
\fill (1,-1) circle (4pt);
\node[right, fill=white] at (4,-3) {$P$};
\draw (-2,1) -- (1,-1) -- (4,-3);
\fill[red] (4,-3) circle (4pt);
\end{tikzpicture}
\end{adjustbox}
\end{document}

答案2

可以使用\useasboundingbox。您需要找到矩形的尺寸才能正确放置图片。

 \documentclass{article}
 \usepackage{tikz}
 \usetikzlibrary{arrows}
 \usepackage{amsmath}
 \usepackage{mathtools}

 \begin{document}
 Let $P = (x,y)$. We need to find $x$ and $y$. We have
 \begin{align*}
 \text{Midpoint} = \biggl( \frac{-2 + x}{2} , \frac{1 + y}{2} \biggr)
 \quad   &\implies \quad
 (1,-1) = \biggl( \frac{-2 + x}{2} , \frac{1 + y}{2} \biggr)\\
 \begin{tikzpicture}[>=latex',scale=0.5] 
 \useasboundingbox (0,0) rectangle (7,3);        
 % Draw grid lines
 \draw[help lines] (-2.5,-3.5) grid (4.5,1.5);
 % Draw x-axis
 \draw[very thick,->] (-3,0) -- (5.5,0)
     node[right] {\large $x$}; 
 % Draw y-axis
 \draw[very thick, ->] (0,-4) -- (0,2.5) 
 node[above] {\large $y$};
 % Special points
 \node[above, fill=white] at (-2,1) {$(-2,1)$};
 \fill (-2,1) circle (4pt);
 \node[right, fill=white] at (1.5,-1) {$(1,-1)$};
 \fill (1,-1) circle (4pt);
 \node[right, fill=white] at (4,-3) {$P$};
 \draw (-2,1) -- (1,-1) -- (4,-3);
 \fill[red] (4,-3) circle (4pt);
 \end{tikzpicture}          &\implies \quad
 \begin{dcases}
     1 = \frac{-2 + x}{2}\\
     -1 = \frac{1 + y}{2}
 \end{dcases}\\
         &\implies \quad
 \begin{dcases}
     2 = -2 + x\\
     -2 = 1 + y
 \end{dcases}
 \quad \implies \quad
 \begin{dcases}
     x = 4\\
     y = -3.
 \end{dcases}
 \end{align*}

 \end{document}

在此处输入图片描述

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