回復 11# litlesweetx 的帖子
在網路上買東西，記得請賣家提供售後服務
\(P\left( x,y \right)\)關於直線L 的對稱點是\(Q\left( x',y' \right)\)，垂足\(P'\left( \frac{x'+x}{2},\frac{y'+y}{2} \right)\)
\(\begin{align}
& \left[ \begin{align}
& x' \\
& y' \\
\end{align} \right]=\left[ \begin{matrix}
\cos 2\theta & \sin 2\theta \\
\sin 2\theta & -\cos 2\theta \\
\end{matrix} \right]\left[ \begin{align}
& x \\
& y \\
\end{align} \right] \\
& \frac{1}{2}\left[ \begin{align}
& x'+x \\
& y'+y \\
\end{align} \right]=\frac{1}{2}\left( \left[ \begin{matrix}
\cos 2\theta & \sin 2\theta \\
\sin 2\theta & -\cos 2\theta \\
\end{matrix} \right]\left[ \begin{align}
& x \\
& y \\
\end{align} \right]+\left[ \begin{align}
& x \\
& y \\
\end{align} \right] \right) \\
& =\frac{1}{2}\left( \left[ \begin{matrix}
\cos 2\theta & \sin 2\theta \\
\sin 2\theta & -\cos 2\theta \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right] \right)\left[ \begin{align}
& x \\
& y \\
\end{align} \right] \\
& =\frac{1}{2}\left( \left[ \begin{matrix}
\cos 2\theta & \sin 2\theta \\
\sin 2\theta & -\cos 2\theta \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right] \right)\left[ \begin{align}
& x \\
& y \\
\end{align} \right] \\
\end{align}\)
