## 利用行列式推導三角形的四心座標公式

$\displaystyle \Delta=\frac{1}{2}\begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1 \end{vmatrix}$

$\Delta=0$，則 $A, B, C$ 三點位於同一直線上，或稱共線。

$\displaystyle \begin{vmatrix} x&y&1\\ x_1&y_1&1\\ x_2&y_2&1 \end{vmatrix}=0$

$\displaystyle \begin{vmatrix} x&y&1\\ a&0&1\\ 0&b&1 \end{vmatrix}=0$

$\displaystyle \frac{x}{a}+\frac{y}{b}=1$

\displaystyle\begin{aligned} \begin{vmatrix} x&y&1\\ x_1&y_1&1\\ x_2&y_2&1 \end{vmatrix}&=\begin{vmatrix} x-x_1&y-y_1&0\\ x_1&y_1&1\\ x_2&y_2&1 \end{vmatrix}=\begin{vmatrix} x-x_1&y-y_1&0\\ x_1-x_2&y_1-y_2&0\\ x_2&y_2&1 \end{vmatrix}\\ &=\begin{vmatrix} x-x_1&y-y_1\\ x_1-x_2&y_1-y_2 \end{vmatrix}=0,\end{aligned}

$\displaystyle \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

$x_1\neq x_2$，直線方程式可表示為

$\displaystyle \begin{vmatrix} x&y&1\\ x_1&y_1&1\\ x_2&y_2&1 \end{vmatrix}=\begin{vmatrix} x-x_1&y-y_1\\ x_1-x_2&y_1-y_2 \end{vmatrix}=(x_1-x_2)\begin{vmatrix} x-x_1&y-y_1\\ 1&m \end{vmatrix}=0$

$\displaystyle y-y_1=m(x-x_1)$

$\displaystyle \begin{vmatrix} x&y&1\\ x_1&y_1&1\\ 1&m&0 \end{vmatrix}=0$

$\displaystyle \begin{vmatrix} x&y&1\\ x_3&y_3&1\\ x_2-x_1&y_2-y_1&0 \end{vmatrix}=0$

$\displaystyle \begin{vmatrix} x&y&1\\ x_3&y_3&1\\ 1&\frac{y_2-y_1}{x_2-x_1}&0 \end{vmatrix}=\frac{1}{x_2-x_1}\begin{vmatrix} x&y&1\\ x_3&y_3&1\\ x_2-x_1&y_2-y_1&0 \end{vmatrix}=0$

$\displaystyle \begin{vmatrix} x&y&1\\ x_3&y_3&1\\ y_2-y_1&x_1-x_2&0 \end{vmatrix}=0$

$\displaystyle \begin{vmatrix} x&y&1\\ x_3&y_3&1\\ 1&-\frac{x_2-x_1}{y_2-y_1}&0 \end{vmatrix}=\frac{1}{y_2-y_1}\begin{vmatrix} x&y&1\\ x_3&y_3&1\\ y_2-y_1&x_1-x_2&0 \end{vmatrix}=0$

\displaystyle\begin{aligned} \begin{vmatrix} x&y&1\\ x_1&y_1&1\\ \frac{x_2+x_3}{2}&\frac{y_2+y_3}{2}&1 \end{vmatrix}&=\frac{1}{2}\begin{vmatrix} x&y&1\\ x_1&y_1&1\\ x_2+x_3&y_2+y_3&2 \end{vmatrix}\\ &=\frac{1}{2}\begin{vmatrix} x&y&1\\ x_1&y_1&1\\ x_1+x_2+x_3&y_1+y_2+y_3&3 \end{vmatrix}=0.\end{aligned}

$\displaystyle \begin{vmatrix} x&y&1\\ x_2&y_2&1\\ x_1+x_2+x_3&y_1+y_2+y_3&3 \end{vmatrix}=0$

$\displaystyle G_x=\frac{1}{3}(x_1+x_2+x_3),~~~G_y=\frac{1}{3}(y_1+y_2+y_3)$

\displaystyle\begin{aligned} \begin{vmatrix} x&y&1\\ x_1&y_1&1\\ \frac{bx_2+cx_3}{b+c}&\frac{by_2+cy_3}{b+c}&1 \end{vmatrix}&=\frac{1}{b+c}\begin{vmatrix} x&y&1\\ x_1&y_1&1\\ bx_2+cx_3&by_2+cy_3&b+c \end{vmatrix}\\ &=\frac{1}{b+c}\begin{vmatrix} x&y&1\\ x_1&y_1&1\\ ax_1+bx_2+cx_3&ay_1+by_2+cy_3&a+b+c \end{vmatrix}=0.\end{aligned}

$\displaystyle \begin{vmatrix} x&y&1\\ x_2&y_2&1\\ ax_1+bx_2+cx_3&ay_1+by_2+cy_3&a+b+c \end{vmatrix}=0$

$\displaystyle I_x=\frac{ax_1+bx_2+cx_3}{a+b+c},~~~I_y=\frac{ay_1+by_2+cy_3}{a+b+c}$

$\displaystyle \begin{vmatrix} x^2+y^2&x&y&1\\ x_1^2+y_1^2&x_1&y_1&1\\ x_2^2+y_2^2&x_2&y_2&1\\ x_3^2+y_3^2&x_3&y_3&1 \end{vmatrix}=0$

$\displaystyle (x^2+y^2)\begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1 \end{vmatrix}-x\begin{vmatrix} x_1^2+y_1^2&y_1&1\\ x_2^2+y_2^2&y_2&1\\ x_3^2+y_3^2&y_3&1 \end{vmatrix}+y\begin{vmatrix} x_1^2+y_1^2&x_1&1\\ x_2^2+y_2^2&x_2&1\\ x_3^2+y_3^2&x_3&1 \end{vmatrix}=\begin{vmatrix} x_1^2+y_1^2&x_1&y_1\\ x_2^2+y_2^2&x_2&y_2\\ x_3^2+y_3^2&x_3&y_3 \end{vmatrix}$

$\displaystyle O_x=\frac{\begin{vmatrix} x_1^2+y_1^2&y_1&1\\ x_2^2+y_2^2&y_2&1\\ x_3^2+y_3^2&y_3&1 \end{vmatrix}}{4\Delta},~~~O_y=\frac{\begin{vmatrix} x_1&x_1^2+y_1^2&1\\ x_2&x_2^2+y_2^2&1\\ x_3&x_3^2+y_3^2&1 \end{vmatrix}}{4\Delta}.$

$\displaystyle \begin{vmatrix} x&y&1\\ x_1&y_1&1\\ y_3-y_2&x_2-x_3&0 \end{vmatrix}=0$

$\displaystyle \begin{vmatrix} x&y&1\\ x_2&y_2&1\\ y_3-y_1&x_1-x_3&0 \end{vmatrix}=0$

\displaystyle\begin{aligned} H_x&=\frac{\begin{vmatrix} x_1x_3-x_1x_2+y_1y_3-y_1y_2&y_3-y_2\\ x_2x_3-x_1x_2+y_2y_3-y_1y_2&y_3-y_1 \end{vmatrix}}{\begin{vmatrix} x_3-x_2&y_3-y_2\\ x_3-x_1&y_3-y_1 \end{vmatrix}}\\ &=\frac{\begin{vmatrix} x_1x_3-x_1x_2+y_1y_3-y_1y_2&y_3-y_2&0\\ x_2x_3-x_1x_2+y_2y_3-y_1y_2&y_3-y_1&0\\ x_1x_2+y_1y_2&y_3&1 \end{vmatrix}}{-\begin{vmatrix} x_1-x_3&y_1-y_3&0\\ x_2-x_3&y_2-y_3&0\\ x_3&y_3&1 \end{vmatrix}}\\ &=-\frac{\begin{vmatrix} x_2x_3+y_2y_3&y_1&1\\ x_3x_1+y_3y_1&y_2&1\\ x_1x_2+y_1y_2&y_3&1 \end{vmatrix}}{\begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1 \end{vmatrix}}=-\frac{\begin{vmatrix} x_2x_3+y_2y_3&y_1&1\\ x_3x_1+y_3y_1&y_2&1\\ x_1x_2+y_1y_2&y_3&1 \end{vmatrix}}{2\Delta}. \end{aligned}

$\displaystyle H_y=-\frac{\begin{vmatrix} x_1&x_2x_3+y_2y_3&1\\ x_2&x_3x_1+y_3y_1&1\\ x_3&x_1x_2+y_1y_2&1 \end{vmatrix}}{2\Delta}$

\displaystyle\begin{aligned} G_x&=\frac{2}{3}O_x+\frac{1}{3}H_x\\ G_y&=\frac{2}{3}O_y+\frac{1}{3}H_y. \end{aligned}

\displaystyle\begin{aligned} 2\Delta(3G_x-2O_x-H_x)&=(x_1+x_2+x_3)\begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1 \end{vmatrix}-\begin{vmatrix} x_1^2+y_1^2&y_1&1\\ x_2^2+y_2^2&y_2&1\\ x_3^2+y_3^2&y_3&1 \end{vmatrix}+\begin{vmatrix} x_2x_3+y_2y_3&y_1&1\\ x_3x_1+y_3y_1&y_2&1\\ x_1x_2+y_1y_2&y_3&1 \end{vmatrix}\\ &=\begin{vmatrix} x_1x_2+x2x_3+x_3x_1+y_2y_3-y_1^2&y_1&1\\ x_1x_2+x2x_3+x_3x_1+y_3y_1-y_2^2&y_2&1\\ x_1x_2+x2x_3+x_3x_1+y_1y_2-y_3^2&y_3&1 \end{vmatrix}\\ &=(x_1x_2+x_2x_3+x_3x_1)\begin{vmatrix} 1&y_1&1\\ 1&y_2&1\\ 1&y_3&1 \end{vmatrix}+\begin{vmatrix} y_2y_3&y_1&1\\ y_3y_1&y_2&1\\ y_1y_2&y_3&1 \end{vmatrix}-\begin{vmatrix} y_1^2&y_1&1\\ y_2^2&y_2&1\\ y_3^2&y_3&1 \end{vmatrix}\\ &=(y_3-y_2)(y_1-y_3)(y_2-y_1)-(y_3-y_2)(y_1-y_3)(y_2-y_1)=0. \end{aligned}

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### 6 Responses to 利用行列式推導三角形的四心座標公式

1. Watt Lin says:

很精彩的文章！
以下這句話看來有不小心打錯一個字
「歐拉 (Leonhard Euler) 證明任意三角形的重心、外心和重心位於一條直線上，稱為歐拉線」
第二個「重心」應該是「垂心」吧！

• ccjou says:

謝謝，已訂正。是該換一付老花眼鏡了。

2. says:

垂心的公式我看不太懂……

3. jerry99980 says:

老師，那如果是空間中的三角形呢？

• ccjou says:
• jerry99980 says:

嗯，我試試看。謝謝老師！