# Find intersection point between two plotted lines

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Mark Sc on 18 Apr 2023
Edited: Star Strider on 18 Apr 2023
Hi all,
I have been trying to find an intersection point between two plotted lines however, I used several functions but it doesnot work.. usually the output is 2*0 empty double matrix ... Actually i need the number of intersection ... ?
here is the code and attached are the data
clearvars
clc;
close all
x = Data(:,1);
y = Data(:,2);
[ymax,idx] = max(y);
Binit = x(1:5) \ y(1:5)
Bymax = x(idx) \ y(idx)
Line_init = x*Binit;
Line_initv = Line_init <= ymax;
Line_ymax = x*Bymax;
Line_ymaxv = Line_ymax <= ymax;
figure
plot(x, y)
hold on
plot([0;x(Line_ymaxv)], [0;Line_ymax(Line_ymaxv)], '-r')
hold off
grid
axis('equal')
xlabel('X')
ylabel('Y')
axis([0 max(x) 0 max(y)])
x1=x(Line_ymaxv)
y1=Line_ymax(Line_ymaxv)
P=InterX([x;y],[x1;y1])
The function i used is from "https://www.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections"

Star Strider on 18 Apr 2023
The code I wrote for your previous post, Plot tangent line on part of the curve is designed as requested to have only one intersection, that being at the origin, since both tangent lines were requested to go through the origin, at least as I understood it.
How else would you want to define the two tangent lines?
Mark Sc on 18 Apr 2023
just one intersect point thats' what I wanna ...
I just wanna a code to let me know the intersection point value
Thank you once again
Star Strider on 18 Apr 2023
Edited: Star Strider on 18 Apr 2023
My pleasure!
The two lines I plotted (only one is shown here) both intersect at the origin, since that is how they were requested and designed. The intersection with the curve is designed to be at ‘ymax’ with the x-coordinate of that intersection being ‘x(idx)’, so nothing further needs to be computed.
% clearvars
% clc;
% close all
format long
Data = 13042×2
1.980321668300000 7.370000000000000 1.982328936500000 7.350000000000000 1.985318034200000 7.380000000000000 1.987310766000000 7.400000000000000 1.990343472900000 7.310000000000000 1.993270790900000 7.510000000000000 1.995263522700000 7.530000000000000 1.997299863700000 7.430000000000000 2.000270790900000 7.510000000000000 2.002187206600000 7.740000000000000
x = Data(:,1);
y = Data(:,2);
[ymax,idx] = max(y);
Binit = x(1:fix(idx/4)) \ y(1:fix(idx/4))
Binit =
3.644815761434447
Bymax = x(idx) \ y(idx)
Bymax =
3.216013064246002
Line_init = x*Binit;
Line_initv = Line_init <= ymax;
Line_ymax = x*Bymax;
Line_ymaxv = Line_ymax <= ymax;
Intersection_x = interp1(Line_init-Line_ymax, x, 0, 'linear','extrap')
Intersection_x =
2.273736754432321e-13
Intersection_y = interp1(x, Line_init, Intersection_x, 'linear','extrap')
Intersection_y =
9.094947017729282e-13
Intersection = [Intersection_x, Intersection_y] % Desired Result
Intersection = 1×2
1.0e-12 * 0.227373675443232 0.909494701772928
figure
plot(x, y)
hold on
plot([0;x(Line_ymaxv)], [0;Line_ymax(Line_ymaxv)], '-r')
plot([0;x(Line_initv)], [0;Line_init(Line_initv)], '-r')
hold off
grid
axis('equal')
xlabel('X')
ylabel('Y')
axis([0 max(x) 0 max(y)])
% x1=x(Line_ymaxv)
% y1=Line_ymax(Line_ymaxv)
% P=InterX([x;y],[x1;y1])
EDIT — (18 Apr 2023 at 20:03)
Added ‘Intersection’ calculation and result using interp1 to calculate the coordinates.
.

Cameron on 18 Apr 2023
Edited: Cameron on 18 Apr 2023
Looks like a stress-strain curve. Depending on the material, you could adjust the variable I named cutoff to 0.3*ymax or whatever you want. This code takes the first intersection of the stress-strain curve and interpolates the value for yield.
clearvars
clc;
close all
x = Data(:,1);
y = Data(:,2);
[ymax,idx] = max(y);
Binit = x(1:5) \ y(1:5);
Bymax = x(idx) \ y(idx);
Line_init = x*Binit;
Line_initv = Line_init <= ymax;
Line_ymax = x*Bymax;
Line_ymaxv = Line_ymax <= ymax;
Line_y = Line_ymax(Line_ymaxv);
trunc_x = x(1:find(Line_ymaxv,1,'last'));
trunc_y = y(1:find(Line_ymaxv,1,'last'));
ii = length(trunc_y);
cutoff = 0.5*ymax; %you can adjust this
SaveMe = [];
while trunc_y(ii) > cutoff
if (Line_y(ii) > trunc_y(ii) && Line_y(ii-1) <= trunc_y(ii-1)) | ...
(Line_y(ii) < trunc_y(ii) && Line_y(ii-1) >= trunc_y(ii-1))
SaveMe = [SaveMe;ii];
end
ii = ii - 1;
end
m1 = polyfit(x(SaveMe(end)-1:SaveMe(end)),y(SaveMe(end)-1:SaveMe(end)),1);
m2 = polyfit(x(SaveMe(end)-1:SaveMe(end)),Line_y(SaveMe(end)-1:SaveMe(end)),1);
x_cross = (m2(2) - m1(2))/(m1(1) - m2(1));
y_cross = m1(1)*x_cross + m1(2);
figure
plot(x, y)
hold on
plot([0;x(Line_ymaxv)], [0;Line_ymax(Line_ymaxv)], '-r')
scatter(x_cross,y_cross,'filled')
grid
axis('equal')
xlabel('X')
ylabel('Y')
axis([0 max(x) 0 max(y)])
hold off
##### 3 CommentsShow 1 older commentHide 1 older comment
Cameron on 18 Apr 2023
The answer is in the code I provided as x_cross and y_cross.
Mark Sc on 18 Apr 2023
Thank you Cameron, but what If I already have my own x and y
x1_y1 as arrays.....
I would appreciate if I can just enter these arrays and got the point of intersection..
Thanks,