已知三点,求三角形面积

mac2022-06-30  24

已经知道三角形三点A(X1,Y1) B(X2,Y2) C(X3,Y3)\[ \vec{AB} = (X2-X1,Y2-Y1) \]

\[ \vec{AC} = (X3-X1,Y3-Y1) \]

\[ ||n|| = \vec{AB} \times \vec{AC} = |\vec{AB}|\cdot|\vec{AB}|*Sin<\vec{AB},\vec{AC}> \]

\[ 因为 |\vec{AB}|*Sin<\vec{AB},\vec{AC}> 为三角形的高 \]

\[ 所以 S_{三角形}= \frac{1}{2} | \vec{AB} \times \vec{AC}| \]

\[= \begin{vmatrix} X2-X1 & Y2-Y1\\ X3-X1 & Y3-Y1 \end{vmatrix} \]

\[ = (X2-X1)(Y3-Y1) * (X3-X1)(Y2-Y1) \]

\[ = X1Y2 + X2Y3 + X3Y1 - X1Y3 - X2Y1 - X3Y2 \]

转载于:https://www.cnblogs.com/--zz/p/11209429.html

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