Determine the intersection Point of the two Diagonals in a quadrilateral

Question

John 2013-5-17

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/76138-determine-the-intersection-point-of-the-two-diagonals-in-a-quadrilateral

评论： Ryan Cho 2017-12-20

Hey people,

is there an easy way of dermining the intersection points of two diagonals in a quadriliteral??

Doing it by hand is quite hard since there are many special cases to consider. maybe there is already a function that helps solving the problem?

With best regards, John

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Ryan Cho 2017-12-20

If you are doing this on a graph get the x values of the coordinates and average them, then get the y values of the coordiantes and averagae them and then there you go (I've only tried this with parralellograms so far bu good luck)

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Answer 1

Image Analyst 2013-5-17

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https://ww2.mathworks.cn/matlabcentral/answers/76138-determine-the-intersection-point-of-the-two-diagonals-in-a-quadrilateral#answer_85778

在 MATLAB Online 中打开

Calculate the equation of the two lines and set them equal, and solve for x, and then plug in x into either equation and get y. Just basic math

m1*x+b1 = m2*x+b2

and so on to solve for x. m is just the slope deltaY/deltaX you get from using your two diagonal points. You can use

coeffs = polyfit(x,y,1); % coeffs(1) = m, coeffs(2) = b

if you want.

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John 2013-5-17

在 MATLAB Online 中打开

Hi Image Analyst, i dont know in advance what pointconnections build the diagonals, so i have to try all 3 kombinations..

i have to add the examnination if the resulting points actually lie in the konvex hull of the four points...i dont know how to implement that right now...

     function [ Steinerpunkt4 ] = Steineropt4( Vektor1, Vektor2, Vektor3, Vektor4 )
B=zeros(2,3);
%Testing, if line from point 1 to 3 and line from 2 to 4 is a   %Steinerpoint

%Derminition of the line-equations

m1=(Vektor1(2)-Vektor3(2))/(Vektor1(1)-Vektor3(1));
b1=Vektor3(2)-m1*Vektor3(1);
m2=(Vektor2(2)-Vektor4(2))/(Vektor2(1)-Vektor4(1));
b2=Vektor4(2)-m2*Vektor4(1);
if isequal(m1,m2)          % in case of parallel lines
     a=1;
else a=0;
end

%to avoid zu high slope so that there is no zero-division

if (( m1 < -(10^(10)) || m1 > 10^(10)) && a==0 )   
        Steinerpunkty=m2*Vektor1(1)+b2;                              
        Steinerpunkt4=[Vektor1(1), Steinerpunkty];
        B(:,1)=Steinerpunkt4;
elseif (( m2 < -(10^(10)) || m2 > 10^(10)) && a==0 )  
        Steinerpunkty=m1*Vektor2(1)+b1;
        Steinerpunkt4=[Vektor2(1), Steinerpunkty];
        B(:,1)=Steinerpunkt4;
elseif a==0
        A=[-m1 ,1; -m2, 1 ];
        b=[b1;b2];
        Steinerpunkt4 = A\b;
        B(:,1)=Steinerpunkt4;
else B(:,1)=[0,0];
end
%Now testing if 1 and 4 and line 2 to 3 form a Steinerpoint
m1=(Vektor1(2)-Vektor4(2))/(Vektor1(1)-Vektor4(1));
b1=Vektor4(2)-m1*Vektor4(1);
m2=(Vektor2(2)-Vektor3(2))/(Vektor2(1)-Vektor3(1));
b2=Vektor3(2)-m2*Vektor3(1);
if isequal(m1,m2)
     a=1;
else a=0;
end
if (( m1 < -(10^(10)) || m1 > 10^(10)) && a==0 )              
        Steinerpunkty=m2*Vektor1(1)+b2;
        Steinerpunkt4=[Vektor1(1), Steinerpunkty];
        B(:,2)=Steinerpunkt4;
elseif (( m2 < -(10^(10)) || m2 > 10^(10)) && a==0 )
        Steinerpunkty=m1*Vektor2(1)+b1;
        Steinerpunkt4=[Vektor2(1), Steinerpunkty];
        B(:,2)=Steinerpunkt4;
elseif a==0
        A=[-m1 ,1; -m2, 1 ];
        b=[b1;b2];
        Steinerpunkt4 = A\b;
        B(:,2)=Steinerpunkt4;
else B(:,2)=[0,0];
end
%No testing if 1 and 2 with line 3 to 4 intersect at a Steinerpoint
m1=(Vektor1(2)-Vektor2(2))/(Vektor1(1)-Vektor2(1));
b1=Vektor2(2)-m1*Vektor2(1);
m2=(Vektor3(2)-Vektor4(2))/(Vektor3(1)-Vektor4(1));
b2=Vektor4(2)-m2*Vektor4(1);
if isequal(m1,m2)
     a=1;
else a=0;
end
if (( m1 < -(10^(10)) || m1 > 10^(10)) && a==0 )         
        Steinerpunkty=m2*Vektor1(1)+b2;                 
        Steinerpunkt4=[Vektor1(1), Steinerpunkty];
        B(:,3)=Steinerpunkt4;
elseif (( m2 < -(10^(10)) || m2 > 10^(10)) && a==0 ) 
        Steinerpunkty=m1*Vektor2(1)+b1;
        Steinerpunkt4=[Vektor2(1), Steinerpunkty];
        B(:,3)=Steinerpunkt4;
elseif a==0 
    A=[-m1 ,1; -m2, 1 ];
    b=[b1;b2];
    Steinerpunkt4 = A\b; 
    B(:,3)=Steinerpunkt4;
else B(:,3)=[0,0];
end
B(:,~any(B))=[];
length(B)
%Making sure that input Vektors/points are not returned
for i=1:length(B)-1
    if  B(1,i)==Vektor1(1) & B(2,i)==Vektor1(2)
        B(:,i)=[]
    end
    if  B(1,i)==Vektor2(1) & B(2,i)==Vektor2(2)
        B(:,i)=[]
    end
    if B(1,i)==Vektor3(1) & B(2,i)==Vektor3(2)
        B(:,i)=[]
    end
    if B(1,i)==Vektor4(1) & B(2,i)==Vektor4(2)
        B(:,i)=[]
    end
i=i+1;
end

end

Image Analyst 2013-5-17

You HAVE to know the order of the points going around the quadrilateral. If you don't know the order, then there are multiple possible quadrilaterals you could build from the same set of points. So, because you must know the order, your diagonals are always points 1 and 3, and points 2 and 4. Always.

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