Finding intersection point of the lines
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Hi I have data sets for two lines. i.e. x1,y1 and x2,y2. So i can plot the lines using these point data sets. I would like to know the point (x,y)where these lines intersect each other. Please note that i have tried both [x,y]=intersections(x1,y1,x2,y2); and [x,y]=curveintersect(x1,y1,x2,y2);
i would appreciate if you can tell me the exact command for this purpose.
Regards
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Andrei Bobrov
2012 年 3 月 5 日
data = rand(20,3);
x1 = sort(data(:,2));
x2 = sort(data(:,3));
y = data(:,1);
pp = interp1(x1,y,'linear','pp');
pp2 = interp1(x2,y,'linear','pp');
xx = xx(max(x1(1),x2(1)) <= xx & min(x1(end),x2(end)) >= xx);
func = @(x)ppval(pp,x)-ppval(pp2,x);
xb = xx([true; diff(func(xx) > 0) ~= 0]);
i1 = hankel(1:2,2:numel(xb));
xout = arrayfun(@(z)fzero(func, xb(i1(:,z))), (1:size(i1,2))' )
mohammed wasiullah
2017 年 4 月 5 日
how to find the intersection between the curve and the straight ?
1 件のコメント
Tan Kah Loon
2017 年 4 月 18 日
Example,y1=x^2+2x+3,y2=2x^2+3x+4 , you have to combine two eq and you get ((x^2+X+1)), type f=[1 1 1] to get the polynomials func and roots (f) for its roots.Next, you have to type your 1st equation into p=[1 2 3], after that, pvals=polyval (p,-0.5) and you will find the 1st intersection. The 2nd intersection use back the same method to find.
Preetham Manjunatha
2022 年 2 月 8 日
Here is the link to find the intersection point of two line segments/lines. A fast two line intersection point finder based on the line parametric space. Finds the intersection point between two lines if it exists or else submits NaN.
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