
Re: [幾何] 三角形座標

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: Sent from JPTT on my Google Pixel 5.
a座標=(x1,y1),da長度d=z1
b座標=(x2,y2),db長度f=z2
c座標=(x3,y3),dc長度e=z3
d座標=(x,y)
以三距離列式: zi^2 = (x-xi)^2 + (y-yi)^2
z1^2 - z2^2 = (2x-x2-x1)(x2-x1) + (2y-y2-y1)(y2-y1)
z2^2 - z3^2 = (2x-x3-x2)(x3-x2) + (2y-y3-y2)(y3-y2)
2(x2-x1)x + 2(y2-y1)y = z1^2 - z2^2 + x2^2 -x1^2 + y2^2 - y1^2
2(x3-x2)x + 2(y3-y2)y = z2^2 - z3^2 + x3^2 -x2^2 + y3^2 - y2^2
解聯立或直接代Cramer's rule
距離合理的話就會有解.
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