How can I find intersection points without mouse?

Hi,

I hope you are keeping safe.

Couls you please help me regarding my matlab code, It is urgent and I would appreciate if you help me. I want to find intersections in this plot without mouse(Using code).

%% Input Parameters
clear all
clc
alpha1=input('Enter positive "alpha1"?');
alpha2=input('Enter positive  "alpha2"?');
beta1=input('Enter positive  "beta1"?');
beta2=input('Enter positive  "beta2"?');
lambda=input('Enter positive  "lambda"?');
eta1=input('Enter positive  "eta1"?');
eta2=input('Enter positive  "eta2"?');
P=input('Enter positive  "P"?');
%% Feasible Area
x = -1:P;m = 0; c = eta2; y = m * x + c;
plot(x, y, 'black')
y = -1:P; m = 0; c = eta1; y = m * x + c;
plot(y, x, 'black')

eta1=min (eta1,P);  eta2=min (eta2,P);
if eta1+eta2<P
    X=[0 0 eta1  eta1];     Y=[0 eta2 eta2 0];
    fill(X,Y,[0.85 0.85 0.85]);
else
X=[0 0 P-eta2 eta1 eta1]; Y=[0 eta2 eta2 P-eta1 0];
fill(X,Y,[0.85 0.85 0.85]); axis([-1 P -1 P])
end

X=[0 0]; Y=[0 P];
line(X,Y,'Color','black')
hold on
line(Y,X,'Color','black')

g = @(x,y) x+y-P;
gp =fimplicit(g,[0 P 0 P], 'black')

f = @(x,y) log((1+alpha1*x./(1+alpha2*y))) -lambda*log((1+beta1*x./(1+beta2*y)));
fp =fimplicit(f,[-1 P -1 P], '-')
hold on
grid on

For alpha1=1 , alpha2=0.2, beta1=1,beta2=2, lambda=2, eta1=3,eta2=4, P=2

Many Thanks in advance

[Jan]

Please remember, that all questions are urgent for the asking persons. Therefore it is polite not to claim, that a question is more urgent than others.

If the answer is important for you, care for asking a clear and unique question. What does "find intersections in this plot without mouse(Using code)" exactly mean? What kind of "intersection" is meant and how would you determine it with the mouse?

[Hossein Alishahi]

Firstly , If you read my request carefully, I have not claimed that my question is more unrgent than others. Secondly, you can run ginput function and then select your desired coordinate from lines by using mouse.

Thanks,

[Jan]

Some members of this forum react allergic to the term "urgent", see e.g. https://www.mathworks.com/matlabcentral/answers/29922-why-your-question-is-not-urgent-or-an-emergency .

You are drawing some objects. I cannot guess between which of these obejcts you want to locate the interection. The more the readers have to guess, the longer takes it to answer.

Answer 1

[Gautam]

Hello Hossein,

If you want to find the point of intersection of any tow lines programmatically, you can do so following the steps below:

1. Create handles to the line objects for which you want to find the intersection point, like:

line1 = line(X,Y,'Color','black');
line2 = line(Y,X,'Color','black');

2. Use the “XData” and “YData” properties of the line objects to find the point of intersection:

interX = line1.XData(line1.XData == line2.XData);>>interY = line1.YData(line1.YData == line2.YData);

3. You can show the point of intersection on the plot using

plot(interX, interY, 'rO', 'MarkerSize',6, 'MarkerFaceColor','red')

Plotting the point of intersection of the lines produces the following graph

You can use the same steps to find the point of intersection of any two lines in your code

[Walter Roberson]

Comparing the XData and YData properties of two lines will generally not work.

X = [0 3];
Y = [1 2];
line1 = line(X,Y,'Color','black');
line2 = line(Y,X,'Color','black');

●

interX = line1.XData(line1.XData == line2.XData);
interY = line1.YData(line1.YData == line2.YData);
plot(interX, interY, 'rO', 'MarkerSize',6, 'MarkerFaceColor','red')

There is no-place that the XData from the one line happens to be exactly the XData from the other line.

The XData property does not interpolate. The result is not the same as

cla
X = 0:.1:3;
Y = linspace(1,2,numel(X));
line1 = line(X,Y,'Color','black');
line2 = line(Y,X,'Color','black');

interX = line1.XData(line1.XData == line2.XData);
interY = line1.YData(line1.YData == line2.YData);
plot(interX, interY, 'rO', 'MarkerSize',6, 'MarkerFaceColor','red')