Let me rephrase your question to be what I think it is, then answer it.
You're given points A = (x1, y1) and B = (x2, y2). You want to find a point Z = (x3, y3) such that AZ is perpendicular to AB, and BZ has length h.
The vector from A to B is v = (x2 - x1, y2 - y1). An easy to calculate perpendicular vector to that one is w = (y2 - y1, x1 - x2). The line crossing through A which is perpendicular to AB is represented by F(s) = A + s*w = (x1 + s*(y2 - y1), y1 + s*(x1 - x2)) as s ranges over the real numbers. So we need to pick a value s such that F(s) is h away from B.
From the Pythagorean theorem, the square of the length from F(s) to B is always going to be the square of the distance from F(s) to A, plus the square of the distance from A to B. From which we get the messy expression that we want:
Now plug that expression for s back into F(s) = (x1 + s*(y2 - y1), y1 + s*(x1 - x2)) and you have your point Z. And the other possible answer is the same distance on the other side.
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让我将您的问题改写为我的想法，然后回答。
给定点
A = (x1, y1)和B = (x2, y2)。您想要找到一个点Z = (x3, y3)，使得AZ垂直于AB，并且BZ> 长度为h。从
A到B的向量是v = (x2 - x1, y2 - y1)。一个易于计算的垂直向量是 w = (y2 - y1, x1 - x2) 。穿过A并垂直于AB的线表示为F(s) = A + s*w = (x1 + s*(y2 - y1) ), y1 + s*(x1 - x2))因为s范围超过实数。因此，我们需要选择一个值s，使得F(s)距离B为h。根据毕达哥拉斯定理，从
F(s)到B的长度的平方始终是到F(s)< 的距离的平方/code> 到A，加上从A到B距离的平方。从中我们得到了我们想要的混乱表达式：现在将
s的表达式插回到F(s) = (x1 + s*(y2 - y1), y1 + s*( x1 - x2))并且你得到了你的观点Z。另一个可能的答案是另一侧的距离相同。Let me rephrase your question to be what I think it is, then answer it.
You're given points
A = (x1, y1)andB = (x2, y2). You want to find a pointZ = (x3, y3)such thatAZis perpendicular toAB, andBZhas lengthh.The vector from
AtoBisv = (x2 - x1, y2 - y1). An easy to calculate perpendicular vector to that one isw = (y2 - y1, x1 - x2). The line crossing throughAwhich is perpendicular toABis represented byF(s) = A + s*w = (x1 + s*(y2 - y1), y1 + s*(x1 - x2))assranges over the real numbers. So we need to pick a valuessuch thatF(s)ishaway fromB.From the Pythagorean theorem, the square of the length from
F(s)toBis always going to be the square of the distance fromF(s)toA, plus the square of the distance fromAtoB. From which we get the messy expression that we want:Now plug that expression for
sback intoF(s) = (x1 + s*(y2 - y1), y1 + s*(x1 - x2))and you have your pointZ. And the other possible answer is the same distance on the other side.