诸位 请问怎么可以求得两条直线的交点坐标
我需要求得两条直线的交点坐标的具体数值我该怎么做呢 请诸位指点一下 可否考虑使用两直线的数学表述，用matlab解方程求出交点！ 回复 2 # 雨人 的帖子
这个也可以 但是有没有什么更简单点的方法
aa=solve('y=x','y-2=-x');
本帖最后由 zhouyang664 于 2011-4-19 22:11 编辑
你看看这个行不行？
function P = InterX(L1,varargin)
%INTERX Intersection of curves
% P = INTERX(L1,L2) returns the intersection points of two curves L1
% and L2. The curves L1,L2 can be either closed or open and are described
% by two-row-matrices, where each row contains its x- and y- coordinates.
% The intersection of groups of curves (e.g. contour lines, multiply
% connected regions etc) can also be computed by separating them with a
% column of NaNs as for example
%
% L= [x11 x12 x13 ... NaN x21 x22 x23 ...;
% y11 y12 y13 ... NaN y21 y22 y23 ...]
%
% P has the same structure as L1 and L2, and its rows correspond to the
% x- and y- coordinates of the intersection points of L1 and L2. If no
% intersections are found, the returned P is empty.
%
% P = INTERX(L1) returns the self-intersection points of L1. To keep
% the code simple, the points at which the curve is tangent to itself are
% not included. P = INTERX(L1,L1) returns all the points of the curve
% together with any self-intersection points.
%
% Example:
% t = linspace(0,2*pi);
% r1 = sin(4*t)+2;x1 = r1.*cos(t); y1 = r1.*sin(t);
% r2 = sin(8*t)+2;x2 = r2.*cos(t); y2 = r2.*sin(t);
% P = InterX(,);
% plot(x1,y1,x2,y2,P(1,:),P(2,:),'ro')
% Author : NS
% Version: 3.0, 21 Sept. 2010
% Two words about the algorithm: Most of the code is self-explanatory.
% The only trick lies in the calculation of C1 and C2. To be brief, this
% is essentially the two-dimensional analog of the condition that needs
% to be satisfied by a function F(x) that has a zero in the interval
% , namely
% F(a)*F(b) <= 0
% C1 and C2 exactly do this for each segment of curves 1 and 2
% respectively. If this condition is satisfied simultaneously for two
% segments then we know that they will cross at some point.
% Each factor of the 'C' arrays is essentially a matrix containing
% the numerators of the signed distances between points of one curve
% and line segments of the other.
%...Argument checks and assignment of L2
error(nargchk(1,2,nargin));
if nargin == 1,
L2 = L1; hF = @lt; %...Avoid the inclusion of common points
else
L2 = varargin{1}; hF = @le;
end
%...Preliminary stuff
x1= L1(1,:)';x2 = L2(1,:);
y1= L1(2,:)';y2 = L2(2,:);
dx1 = diff(x1); dy1 = diff(y1);
dx2 = diff(x2); dy2 = diff(y2);
%...Determine 'signed distances'
S1 = dx1.*y1(1:end-1) - dy1.*x1(1:end-1);
S2 = dx2.*y2(1:end-1) - dy2.*x2(1:end-1);
C1 = feval(hF,D(bsxfun(@times,dx1,y2)-bsxfun(@times,dy1,x2),S1),0);
C2 = feval(hF,D((bsxfun(@times,y1,dx2)-bsxfun(@times,x1,dy2))',S2'),0)';
%...Obtain the segments where an intersection is expected
= find(C1 & C2);
if isempty(i),P = zeros(2,0);return; end;
%...Transpose and prepare for output
i=i'; dx2=dx2'; dy2=dy2'; S2 = S2';
L = dy2(j).*dx1(i) - dy1(i).*dx2(j);
i = i(L~=0); j=j(L~=0); L=L(L~=0);%...Avoid divisions by 0
%...Solve system of eqs to get the common points
P = unique([dx2(j).*S1(i) - dx1(i).*S2(j), ...
dy2(j).*S1(i) - dy1(i).*S2(j)]./,'rows')';
function u = D(x,y)
u = bsxfun(@minus,x(:,1:end-1),y).*bsxfun(@minus,x(:,2:end),y);
end
end
回复 5 # zhouyang664 的帖子
谢谢诸位的帮助
回复 6 # 尚慧娟 的帖子
楼主解决了问题，可否分享下自己的方法！ 楼主分享啊~~~~~ 二元一次方程组 你懂的 用参数方程表达直线，然后形成方程组，解方程组，程序十分简单！ 回复 1 # 尚慧娟 的帖子
使两条直线的方程相减等于0，然后求解方程 回复 11 # yxlnbu 的帖子
谢谢诸位了 问题已经解决了
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