回覆 4# mojary 的帖子
填充10.
已知的行列式相加，透過補項，可以寫出 \( \triangle ABC \) (有號)面積
\( \begin{vmatrix}a_{1} & a_{2}\\
b_{1} & b_{2}
\end{vmatrix}+\begin{vmatrix}b_{1} & b_{2}\\
c_{1} & c_{2}
\end{vmatrix}+\begin{vmatrix}c_{1} & c_{2}\\
d_{1} & d_{2}
\end{vmatrix}+\begin{vmatrix}d_{1} & d_{2}\\
a_{1} & a_{2}
\end{vmatrix} \)
\( = \begin{vmatrix}a_{1} & a_{2}\\
b_{1} & b_{2}
\end{vmatrix}+\begin{vmatrix}b_{1} & b_{2}\\
c_{1} & c_{2}
\end{vmatrix}+\left(\begin{vmatrix}c_{1} & c_{2}\\
a_{1} & a_{2}
\end{vmatrix}+\begin{vmatrix}a_{1} & a_{2}\\
c_{1} & c_{2}
\end{vmatrix}\right)+\begin{vmatrix}c_{1} & c_{2}\\
d_{1} & d_{2}
\end{vmatrix}+\begin{vmatrix}d_{1} & d_{2}\\
a_{1} & a_{2}
\end{vmatrix} \)
\( = \left(\begin{vmatrix}a_{1} & a_{2}\\
b_{1} & b_{2}
\end{vmatrix}+\begin{vmatrix}b_{1} & b_{2}\\
c_{1} & c_{2}
\end{vmatrix}+\begin{vmatrix}c_{1} & c_{2}\\
a_{1} & a_{2}
\end{vmatrix}\right)+\left(\begin{vmatrix}a_{1} & a_{2}\\
c_{1} & c_{2}
\end{vmatrix}+\begin{vmatrix}c_{1} & c_{2}\\
d_{1} & d_{2}
\end{vmatrix}+\begin{vmatrix}d_{1} & d_{2}\\
a_{1} & a_{2}
\end{vmatrix}\right)=2\triangle ABC + 2\triangle ACD \)
ABC、ACD 順逆相反，上式的有號面積，一個正、一個負
再從向量的式子，換成圖形上長度關係(分點公式、係數積)
會有 \( |\triangle ACD| = 5 \times \frac35 |\triangle ABC| = 2 |\triangle ABC| \)
故 \( 4|\triangle ABC| = 1 \Rightarrow |\triangle ABC| = \frac14 \)
