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How can I find a intersection of four ponits?

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Yoon Chang Jeon
Yoon Chang Jeon 2017-11-18
评论： Yoon Chang Jeon 2017-11-19
I know
A(n)=[x(n),y(n)],
A(n+1)=[x(n+1),y(n+1)]
A(m)=[x(m),y(m)],
A(m+1)=[x(m+1),y(m+1)]
I want to find a intersection point of Line A(n)-A(n+1) and A(m)-A(m+1) The x-coordinate of point(x,y) between x(n), x(n+1), x(m),and x(m+1) The y-coordinate of point(x,y) between y(n), y(n+1), y(m),and y(m+1)
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Walter Roberson
Walter Roberson 2017-11-18
Are you sure that the last two knowns are both A(m)? Perhaps the last one is to be A(m+1) ?

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回答（2 个）

David Goodmanson
David Goodmanson 2017-11-18
编辑：David Goodmanson 2017-11-18
Hello Yoon, lots of possible expressions for this. Here is one that is pretty symmetric in the points. Let
a = [x(n),y(n)]'
b = [x(n+1),y(n+1)]'
c = [x(m),y(m)]'
d = [x(m+1),y(m+1)]'
so, four column vectors. then
M = [b-a,d-c]
p = (1/4)*(a+b+c+d + M*[1 0;0 -1]*(M\(c+d-a-b)))
Roger Stafford
Roger Stafford 2017-11-18
If A = [x1;y1] and B = [x2;y2] are coordinates of two points as column vectors on one of the lines with C = [x3;y3] and D = [x4;y4] as column vectors for two points on the other line, then the point P = [x;y] of the lines' intersection is:
TS = [B-A,C-D]\[C-A];
P = A+(B-A)*TS(1);
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Yoon Chang Jeon
Yoon Chang Jeon 2017-11-19
Thanks! But It's result over range (x1 - x2) (x3 - x4), (y1 - y2) (y3 - y4). How can I find a result inside range?

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