这个公式是怎么推导出来的?在线等 200
已知直角坐标系上不在同一直线上三点A(x1,y1)B(x2,y2)C(x3,y3)S△ABC=1/2*|(x2-x1)(y3-y1)-(x3-x1)(y2-y1)|还有任...
已知直角坐标系上不在同一直线上三点A(x1,y1)B(x2,y2)C(x3,y3)
S△ABC=1/2 * |(x2-x1)(y3-y1)-(x3-x1)(y2-y1)|
还有任意n边形也都是类似的公式,怎么推导的? 展开
S△ABC=1/2 * |(x2-x1)(y3-y1)-(x3-x1)(y2-y1)|
还有任意n边形也都是类似的公式,怎么推导的? 展开
6个回答
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通过C作AB的平行线,此直线方程: (y-y3)/(x-x3) = (y2-y1)/(x2-x1)
作B垂直于X轴直线,交于平行线交点C', C' X坐标为X2, 带入直线方程得到Y坐标y3+ [(x2-x3)(y2-y1)/(x2-x1)];
点A到BC'的距离为(x2-x1), BC'长度:y3+ [(x2-x3)(y2-y1)/(x2-x1)] - y2
S ABC' = 1/2 * (x2-x1) * [y3+ [(x2-x3)(y2-y1)/(x2-x1)] - y2]
= 1/2 * [(y3-y2)(x2-x1) + (x2-x3)(y2-y1)]
= 1/2 * [(x2-x1)(y3-y2) - (x3-x2)(y2-y1)]
因为三个点有互换性,将x1,y1 与x2,y2互换,
上式 = 1/2 * [(x1-x2)(y3-y1) - (x3-x1)(y1-y2)]
= 1/2 * [ -(x2-x1)(y3-y1) + (x3-x1)(y2-y1)] (恒为正)
= 1/2 * | (x2-x1)(y3-y1) - (x3-x1)(y2-y1)|
因为直线平行于AB,所以SABC = S ABC' ;得证
作B垂直于X轴直线,交于平行线交点C', C' X坐标为X2, 带入直线方程得到Y坐标y3+ [(x2-x3)(y2-y1)/(x2-x1)];
点A到BC'的距离为(x2-x1), BC'长度:y3+ [(x2-x3)(y2-y1)/(x2-x1)] - y2
S ABC' = 1/2 * (x2-x1) * [y3+ [(x2-x3)(y2-y1)/(x2-x1)] - y2]
= 1/2 * [(y3-y2)(x2-x1) + (x2-x3)(y2-y1)]
= 1/2 * [(x2-x1)(y3-y2) - (x3-x2)(y2-y1)]
因为三个点有互换性,将x1,y1 与x2,y2互换,
上式 = 1/2 * [(x1-x2)(y3-y1) - (x3-x1)(y1-y2)]
= 1/2 * [ -(x2-x1)(y3-y1) + (x3-x1)(y2-y1)] (恒为正)
= 1/2 * | (x2-x1)(y3-y1) - (x3-x1)(y2-y1)|
因为直线平行于AB,所以SABC = S ABC' ;得证
追问
那任意n边形呢?
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正多边形可以拆成多个 三角形来推啊
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通过C作AB的平行线,此直线方程: (y-y3)/(x-x3) = (y2-y1)/(x2-x1)
作B垂直于X轴直线,交于平行线交点C', C' X坐标为X2, 带入直线方程得到Y坐标y3+ [(x2-x3)(y2-y1)/(x2-x1)];
点A到BC'的距离为(x2-x1), BC'长度:y3+ [(x2-x3)(y2-y1)/(x2-x1)] - y2
S ABC' = 1/2 * (x2-x1) * [y3+ [(x2-x3)(y2-y1)/(x2-x1)] - y2]
= 1/2 * [(y3-y2)(x2-x1) + (x2-x3)(y2-y1)]
= 1/2 * [(x2-x1)(y3-y2) - (x3-x2)(y2-y1)]
因为三个点有互换性,将x1,y1 与x2,y2互换,
上式 = 1/2 * [(x1-x2)(y3-y1) - (x3-x1)(y1-y2)]
= 1/2 * [ -(x2-x1)(y3-y1) + (x3-x1)(y2-y1)] (恒为正)
= 1/2 * | (x2-x1)(y3-y1) - (x3-x1)(y2-y1)|
作B垂直于X轴直线,交于平行线交点C', C' X坐标为X2, 带入直线方程得到Y坐标y3+ [(x2-x3)(y2-y1)/(x2-x1)];
点A到BC'的距离为(x2-x1), BC'长度:y3+ [(x2-x3)(y2-y1)/(x2-x1)] - y2
S ABC' = 1/2 * (x2-x1) * [y3+ [(x2-x3)(y2-y1)/(x2-x1)] - y2]
= 1/2 * [(y3-y2)(x2-x1) + (x2-x3)(y2-y1)]
= 1/2 * [(x2-x1)(y3-y2) - (x3-x2)(y2-y1)]
因为三个点有互换性,将x1,y1 与x2,y2互换,
上式 = 1/2 * [(x1-x2)(y3-y1) - (x3-x1)(y1-y2)]
= 1/2 * [ -(x2-x1)(y3-y1) + (x3-x1)(y2-y1)] (恒为正)
= 1/2 * | (x2-x1)(y3-y1) - (x3-x1)(y2-y1)|
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通过C作AB的平行线,此直线方程: (y-y3)
/(x-x3) = (y2-y1)/(x2-x1)作B垂直于X轴直线,交于平行线交点C',
C' X坐标为X2, 带入直线方程得到Y坐标y
3+ [(x2-x3)(y2-y1)/(x2-x1)];点A到BC'的距离为(x2-x1), BC'长度:y3+
[(x2-x3)(y2-y1)/(x2-x1)] - y2S ABC' = 1/2 * (x2-x1) * [y3+ [(x2-x3)(y2
-y1)/(x2-x1)] - y2]= 1/2 * [(y3-y2)(x2-x1) + (x2-x3)(y2-y1)]= 1/2 * [(x2-x1)(y3-y2) - (x3-x2)(y2-y1)]因为三个点有互换性,将x1,y1 与x2,y2
互换,上式 = 1/2 * [(x1-x2)(y3-y1) - (x3-x1)(y1
-y2)]= 1/2 * [ -(x2-x1)(y3-y1) + (x3-x1)(y2-y1)
] (恒为正)= 1/2 * | (x2-x1)(y3-y1) - (x3-x1)(y2-y1
)|因为直线平行于AB,所以SABC = S ABC
' ;得证
还可以用水平面的行列式来做,这是大学的内容,在此不做描述。
/(x-x3) = (y2-y1)/(x2-x1)作B垂直于X轴直线,交于平行线交点C',
C' X坐标为X2, 带入直线方程得到Y坐标y
3+ [(x2-x3)(y2-y1)/(x2-x1)];点A到BC'的距离为(x2-x1), BC'长度:y3+
[(x2-x3)(y2-y1)/(x2-x1)] - y2S ABC' = 1/2 * (x2-x1) * [y3+ [(x2-x3)(y2
-y1)/(x2-x1)] - y2]= 1/2 * [(y3-y2)(x2-x1) + (x2-x3)(y2-y1)]= 1/2 * [(x2-x1)(y3-y2) - (x3-x2)(y2-y1)]因为三个点有互换性,将x1,y1 与x2,y2
互换,上式 = 1/2 * [(x1-x2)(y3-y1) - (x3-x1)(y1
-y2)]= 1/2 * [ -(x2-x1)(y3-y1) + (x3-x1)(y2-y1)
] (恒为正)= 1/2 * | (x2-x1)(y3-y1) - (x3-x1)(y2-y1
)|因为直线平行于AB,所以SABC = S ABC
' ;得证
还可以用水平面的行列式来做,这是大学的内容,在此不做描述。
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边长a=根号下(x1-x2)^2+(y1-y2)^2
即a^2=(x1-x2)^2+(y1-y2)^2
分别以A、B点为圆心,a为半径画圆
(x-x1)^2+(y-y1)^2=a^2
(x-x2)^2+(y-y2)^2=a^2
联立求交点,两个方程两个未知数,可解得两个正确解
方法简单 就是解方程麻烦
即a^2=(x1-x2)^2+(y1-y2)^2
分别以A、B点为圆心,a为半径画圆
(x-x1)^2+(y-y1)^2=a^2
(x-x2)^2+(y-y2)^2=a^2
联立求交点,两个方程两个未知数,可解得两个正确解
方法简单 就是解方程麻烦
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