세 점이 이루는 삼각형의 넓이

  

Choose the vertices of the triangle $x_1, x_2, x_3$ to be $x_1<x_2<x_3$, then

 \begin{align*} m_1 = \frac{y_2-y_2}{x_2-x_1}, m_2 = \frac{y_3-y_2}{x_3-x_2}, m_3 = \frac{y_3-y_1}{x_3-x_1}.\end{align*}

 

The area of the triangle $ A $ is

\begin{align*} A =& \int_{x_1}^{x_2}|[m_1(x-x_1)+y_1]-[m_3(x-x_1)+y_1]|dx \\&+ \int_{x_2}^{x_3}|[m_2(x-x_3)+y_3]-[m_3(x-x_3)+y_3]|dx \\=&\int_{x_1}^{x_2}|(m_1-m_3)(x-x_1)|dx\\&+ \int_{x_2}^{x_3}|(m_2-m_3)(x-x_3)|dx \\=& \int_{0}^{x_2-x_1}|(m_1-m_3)x|dx+ \int_{x_2-x_3}^{0}|(m_2-m_3)x|dx \\=& |m_1-m_3|\frac{1}{2}(x_2-x_1)^2+|m_2-m_3|\frac{1}{2}(x_2-x_3)^2.\end{align*}

 

If $m_1<m_3$, then $m_2>m_3$. $\Rightarrow m_1-m_3<0, m_2-m_3>0$

If $m_1>m_3$, then $m_2<m_3$. $\Rightarrow m_1=m_3>0, m_2-m_3<0$

 

So 

\begin{align*} A =& \frac{1}{2}\left|(m_1-m_3)(x_2-x_1)^2-(m_2-m_3)(x_2-x_3)^2\right| \\ =&\frac{1}{2}\left|(y_2-y_1)(x_2-x_1)-\frac{y_3-y_1}{x_3-x_2}(x_2-x_1)^2 \right.\\ &\left.-(y_3-y_2)(x_3-x_2)+\frac{y_3-y_1}{x_3-x_1}(x_3-x_2)^2\right| \\ =& \frac{1}{2}\left|(y_2-y_1)(x_2-x_1)-(y_3-y_2)(x_3-x_2)\right.\\ &\left.+\frac{y_3-y_1}{x_3-x_1}(x_3-x_2-x_2+x_1)(x_3-x_2+x_2-x_1)\right| \\ =& \frac{1}{2}\left|(y_2-y_1)(x_2-x_1)-(y_3-y_2)(x_3-x_2)+(y_3-y_1)(x_3+x_1)\right| \\ =& \frac{1}{2}\left|-x_1y_2+x_2y_1-x_2y_3+x_3y_2-x_3y_1+x_1y_3\right| \\ =& \frac{1}{2}\left|\begin{vmatrix} x_1 & x_2 \\ y_1 & y_2 \end{vmatrix}+\begin{vmatrix} x_2 & x_3 \\ y_2 & y_3 \end{vmatrix} + \begin{vmatrix} x_3 & x_1 \\ y_3 & y_1 \end{vmatrix}\right| \\ =& \frac{1}{2}\left|\begin{vmatrix} 1 & 1 & 1 \\ x_3 & x_1 & x_2 \\ y_3 & y_1 & y_2 \end{vmatrix}\right|\end{align*}

 

Although the values of the x-axis of two vertuces are same, you use this formula by taking the limit.