在 pgfplots 圖表上複製軸標籤的精確定位

Question

|x| + |y| = 6我有和的圖形y = x - 2，並且y = x + 2在不同的笛卡爾平面上。一個是用 繪製的pgfplots，另一個是僅用 繪製的TikZ。我喜歡使用 的軸標籤的位置pgfplots。如何僅使用 -x和y-獲得軸標籤的相同位置TikZ？

\documentclass{amsart}
\usepackage{amsmath}


\usepackage{tikz}
\usetikzlibrary{calc,intersections}



\usepackage{pgfplots}
\pgfplotsset{compat=1.11}


\begin{document}


\begin{tikzpicture}


\begin{axis}[width=3in, height=3in, axis lines=middle, clip=false,
    axis lines=middle, clip=false,
    xmin=-12,xmax=12,
    ymin=-12,ymax=12,
    restrict y to domain=-12:12,
    xtick={\empty}, ytick={\empty},
    axis line style={latex-latex},
    xlabel=\textit{x},ylabel=\textit{y},
    axis line style={shorten >=-12.5pt, shorten <=-12.5pt},
    xlabel style={at={(ticklabel* cs:1)}, xshift=12.5pt, anchor=north west},
    ylabel style={at={(ticklabel* cs:1)}, yshift=12.5pt, anchor=south west}
]

\addplot[samples=2, domain=0:6] {x - 6};
\addplot[samples=2, domain=-6:0] {x + 6};
\addplot[samples=2, domain=0:6] {-x + 6};
\addplot[samples=2, domain=-6:0] {-x - 6};
\addplot[samples=2, latex-latex, domain=-10:10] {x - 2};
\addplot[samples=2, latex-latex, domain=-10:10] {x + 2};

%The intersections between the square |x| + |y| = 6 and two parallel lines are located.
\coordinate (A) at (-4,-2);
\coordinate (B) at (-2,-4);
\coordinate (C) at (4,2);
\coordinate (D) at (2,4);


\coordinate (P_1) at (-3,3);
\coordinate (P_2) at (0,6);

\coordinate (Q_1) at (-9,-7);
\coordinate (Q_2) at (0,2);
\end{axis}


%A "pin" is drawn between the label for the graph of |x| + |y| = 6 and a point on the graph.
\draw[draw=gray, shorten <=1mm, shorten >=1mm] (P_1) -- ($(P_1)!0.75cm!90:(P_2)$);
\node[anchor=east, inner sep=0, font=\footnotesize] at ($(P_1)!0.75cm!90:(P_2)$) {\makebox[0pt][r]{$\vert x \vert + \vert y \vert = 6$}};

%A "pin" is drawn between the label for the graph of y = x + 2 and a point on the graph.
\draw[draw=gray, shorten <=1mm, shorten >=1mm] (Q_1) -- ($(Q_1)!0.75cm!90:(Q_2)$);
\node[anchor=east, inner sep=0, font=\footnotesize] at ($(Q_1)!0.75cm!90:(Q_2)$) {\makebox[0pt][r]{$y = x - 2$}};


\end{tikzpicture}

\begin{tikzpicture}

%On the Cartesian plane, the solution set to |x| + |y| \leq 6 and $|x - y| \geq 2$ is plotted.

\draw[latex-latex] ($({(1/3)*(-10)}, 0) +(-12.5pt,0)$) -- ($({(1/3)*10}, 0) +(12.5pt,0)$);
\draw[latex-latex] ($(0,{(1/3)*(-12)}) +(0,-12.5pt)$) -- ($(0,{(1/3)*12}) +(0,12.5pt)$);
\node [anchor=north west] at ($({(1/3)*10}, 0) +(0.15,0) +(12.5pt,0)$){\textit{x}};
\node [anchor=south west] at ($(0,{(1/3)*12}) +(0,0.15) +(0,12.5pt)$){\textit{y}};


\draw (0,{(1/3)*(-6)}) -- ({(1/3)*6},0) -- (0,{(1/3)*6}) -- ({(1/3)*(-6)},0) -- cycle;

\draw [latex-latex] ({(1/3)*(-10)}, {(1/3)*(-12)}) -- ({(1/3)*10}, {(1/3)*8});
\draw [latex-latex] ({(1/3)*(-10)}, {(1/3)*(-8)}) -- ({(1/3)*10}, {(1/3)*12});


\end{tikzpicture}


\end{document}