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如下图，C 点是以「交点」的方式求得的，并不知其确切坐标，欲作原点(0,0) 关于 点 C 对称的点Q，也即是在 x 轴上找一点Q，令OC=CQ，求问如何引用『线短OC之长』？
```lua
$$\begin{tikzpicture}[scale=0.7]
\draw[-latex] (-7,0) -- (12,0) node [above]{$x$};
\draw[-latex] (0,-7) -- (0,8) node [right]{$y$};
\draw plot[samples=400,domain=0:15](\x,{2*sqrt(\x)}) ;
\draw[name path=a,yscale=-1] plot[samples=400,domain=0:10](\x,{2*sqrt(\x)}) node[right]{$y^2=4{x}$};
\coordinate(A) at(1,2);
\coordinate(B) at(14.66,7.66);
\draw[name path=f](A)--(B);
\coordinate (Q) at ($(A)!-7cm!90:(B)$);
\draw [name path=b](Q)--(A);
\draw[name path=o] (-7,0) -- (12,0) ;
\draw[name path=t] (0,-7) -- (0,8);
\fill [name intersections={of=a and b, by={D}}] (D) circle (4pt) node at (3.6,-3) {$D$};
\draw[name path=c,domain=-7:4] plot(\x,{(-2*\x-6.69)/3.66});
\draw[name path=d,domain=-7:15] plot(\x,{(2*\x+29.31)/7.66}) ;
\fill [name intersections={of=c and d, by={P}}] (P) circle (4pt) node[above] {$P$};
\fill [name intersections={of=o and b, by={C}}] (C) circle (4pt) node at (2,.6) {$C$};
\fill (A) circle (4pt) node at (1,2.5) {$A$};
\fill (B) circle (4pt) node at (14,8) {$B$};
\fill (0,0) circle (4pt) node at (-0.5,-0.5) {$O$};
\draw[name path=g](P)--(A);
\coordinate (W) at ($(A)!-5.5cm!(B)$);
\draw [name path=w](A)--(W);
\fill [name intersections={of=o and w, by={I}}] (I) circle (4pt) node[below] {$I$};
\end{tikzpicture}$$
```
4 回答
0
加载`tkz-euclide`包，输入
```
\coordinate (O) at (0,0);
\tkzDefPointBy[symmetry = center C](O)
\tkzGetPoint{E}
\fill (E) circle(3pt);
```
0
```
\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
\draw[step=1cm](0,0)grid(2,4);
\coordinate (O) at (0,0);
\coordinate (C) at (1,2);
\coordinate (Q) at ($(C)!-1!(O)$);
\draw[fill=white](O)circle(0.5mm)node[below]{$O$}(C)circle(0.5mm)node[below right]{$C$}(Q)circle(0.5mm)node[above]{$Q$};
\end{tikzpicture}
\end{document}`
0
```c
\begin{tikzpicture}
\def\pointradius{1.2pt}
\draw[-latex,name path=x] (-6,0) -- (9,0) node [above]{$x$};%x轴
\draw[-latex] (0,-6) -- (0,6) node [right]{$y$};%y轴
\coordinate (O) at(0,0);
\fill (O) circle [radius=\pointradius] node[below left]{$O$};
\draw[name path=a]%曲线
plot[variable=\t,samples=400,domain=-6:6]({(\t)^2/4},\t)
coordinate (B);
\coordinate (A) at({(2)^2/4},2);
\fill (B) circle [radius=\pointradius] node[above]{$B$}
(A) circle [radius=\pointradius] node[above]{$A$};
\node[right]at([yscale=-1]B){$y^2=4x$};
\path [overlay,name path=AD](A)--($(A)!1!-90:(B)$);
\filldraw [name intersections={of=AD and a, by=D}]
(A)--(D) circle [radius=\pointradius] node [below] {$D$};
\filldraw [name intersections={of=AD and x, by=C}]
(C) circle [radius=\pointradius] node [above right] {$C$};
\fill ($(O)!2!(C)$) circle [radius=\pointradius]
node[above]{$Q$}
coordinate (Q);
%计算切线的倾斜角度，用 y=2*sqrt(x), y=-2*sqrt(x) 的导数
\path (B);
\pgfgetlastxy{\macrox}{\macroy}
\pgfmathparse{atan(1/sqrt(\macrox/28.45274))}%1cm = 28.45274pt
\let\tBangle\pgfmathresult
\path [overlay,name path=BP] (B)--++(\tBangle+180:50);
\path (D);
\pgfgetlastxy{\macrox}{\macroy}
\pgfmathparse{atan(-1/sqrt(\macrox/28.45274))}%1cm = 28.45274pt
\let\tDangle\pgfmathresult
\path[overlay,name path=DP] (D)--++(\tDangle+180:50);
\filldraw [name intersections={of=DP and BP, by=P}]
(D)--(P) (B)--(P) circle [radius=\pointradius] node [above] {$P$};
\path[overlay,name path=AB] (A)--($(B)!10!(A)$);
\filldraw [name intersections={of=x and AB, by=I}]
(B)--(I) circle [radius=\pointradius] node [above] {$I$};
\end{tikzpicture}
```
效果是
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