How to find the coordinates of a point perpendicular to a line (knowing the distance)?

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ARGY B 2019-10-9

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编辑： Adam Danz 2021-5-20

I have a line made of two points A(xA,yA) and B(xB,yB) and I know a point C(xC,yC) belonging to this line (somewhere between A and B).

I want to find the coordinates of point D(xD,yD), knowing that the (CD) is perpendicular to (AB) and knowing also that the part CD is equal to a number (eg 3)

Are you aware of a function already doing it?

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Answer 1

Adam Danz 2019-10-9

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https://ww2.mathworks.cn/matlabcentral/answers/484422-how-to-find-the-coordinates-of-a-point-perpendicular-to-a-line-knowing-the-distance#answer_395533

编辑：Adam Danz 2019-10-9

在 MATLAB Online 中打开

Here's a tutorial with comments so you can follow what's going on. The end points of the perpendicular segment ("D" in your photo) are stored in variables x and y. See note at the end of this answer regarding vertical and horizontal lines.

Set the coordinates for points A and B at the top of the code. Then set the 1/2 length of the perpendicular line.

% Enter coordinates of points A and B and  
% the 1/2 length of the perpendicular segment
A = [3 3]; %[x,y]
B = [12,8]; %[x,y]
Clen = 3;  %length of line CD (1/2 of the full perpendicular line)
%% Do the math
% Get slope and y int of line AB
slope = (B(2)-A(2)) / (B(1)-A(1)); 
yint = B(2) - slope*B(1); 
% Choose a point C along line AB at half distance
C(1) = range([B(1),A(1)])/2+min([A(1),B(1)]); 
C(2) = slope * C*(1) + yint; 
% Get slope and y int of line perpendicular to AB at point C
perpSlope = -1/slope; 
perpYint = C(2) - perpSlope*C(1); 
% Find the end points of the perpendicular line with length Clen*2
x = C(1) + (Clen*sqrt(1/(1+perpSlope^2)))*[-1,1]; 
y = C(2) + (perpSlope*Clen*sqrt(1/(1+perpSlope^2)))*[-1,1]; 
%% Plot results
figure()
% Draw line AB
p1 = plot([A(1),B(1)], [A(2),B(2)], 'k-o','LineWidth',2,'DisplayName','AB'); 
hold on
axis equal 
grid on
axlim = xlim(gca)+[-1,1]*range(xlim(gca))*1.1; 
xlim(axlim)
ylim(axlim)
% Draw a references line parallel to AB
rl = refline(slope,yint); 
set(rl, 'LineStyle', '--', 'Color', [.5 .5 .5])
% Draw a references line perpendicular to AB at point C
rl = refline(perpSlope,perpYint); 
set(rl, 'LineStyle', '--', 'Color', [.5 .5 .5])
xlim(axlim) %refline() changes limits so we set them again :(
ylim(axlim)
% Add point C
p2 = plot(C(1),C(2), 'r*','DisplayName','C');
% Add the perp segment
p3 = plot(x,y,'b-o','LineWidth',2,'DisplayName','perp AB');
legend([p1,p2,p3])

Note that if line AB is perfectly horizontal, you'll run into +|- inf issues with the perpendicular slope. If AB is perfectly vertical, my calculation of C will fail (it will return a NaN due to the inf slope of AB) but you mentioned that you already have C so that shouldn't be an issue.

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John D'Errico 2019-10-9

编辑：John D'Errico 2019-10-9

在 MATLAB Online 中打开

Accept Adam's excellent answer. However, let me point out a slight variation on what he did.

A = [3 3]; %[x,y]
B = [12,8]; %[x,y]
Clen = 3; % distance off the line AB that C will lie
% Call AB the vector that points in the direction
% from A to B.
AB = B - A;
% Normalize AB to have unit length
AB = AB/norm(AB);
% compute the perpendicular vector to the line
% because AB had unit norm, so will ABperp
ABperp = AB*[0 -1;1 0];
% midpoint between A and B
ABmid = (A + B)/2;
% Compute new points C and D, each at a ditance
% Clen off the line. Note that since ABperp is
% a vector with unit eEuclidean norm, if I
% multiply it by Clen, then it has length Clen.
C = ABmid + Clen*ABperp;
D = ABmid - Clen*ABperp;
% plot them all
plot([A(1);B(1)],[A(2);B(2)],'-bo',[C(1);D(1)],[C(2);D(2)],'-rs')
% axis equal is important because it ensures the lines appear
% mutually perpendicular. If the axes had different units
% along the axes, then the lines would look skewed.
axis equal

Note the trick of multiplying AB by [0 -1;1 0] to get a perpendicular vector. You can think of this as a rotation matrix based on sines and cosines, but since the rotation angle is a 90 degree rotation, the matrix looks quite simple.

The nice thing about an approach like what I've shown is you never need to compute explicit slopes of lines. Could a slope computation fail? YES! What if A and B happened to be perfectly horizontal or perfectly vertical? In that case one of the slopes you will compute will be zero, and the other infinite. Bad things will happen as a consequence. But the solution I show will never fail due to such a degeneracy, because I worked purely in terms of vectors. This is an important point to look for in the future, as it can be hugely important in computational geometry code.

One other important point to see is that my solution (as well as Adam's) will fail in one circumstance: where A and B are coincident points. So if the distance between points A and B is zero, then anything you do must fail, since then the vector between the points is indeterminate, as is the perpendicular vector.

Andrew Bliss 2021-1-12

在 MATLAB Online 中打开

I know it's probably not the right place to post this, but I found this post when searching for how to do this in R. John's answer got me there and I thought I would share my translation to R in case anyone else comes along later.

# R code version of John's answer
A = c(3, 3) #[x,y]]
B = c(12,8) #[x,y]
Clen = 3; # distance off the line AB that C will lie
# Call AB the vector that points in the direction
# from A to B.
AB = B - A;
# Normalize AB to have unit length
AB = AB/max(svd(AB)$d)
# compute the perpendicular vector to the line
# because AB had unit norm, so will ABperp
ABperp = AB%*%matrix(c(0,-1,1,0),nrow=2,byrow=TRUE)
# midpoint between A and B
ABmid = (A + B)/2;
# Compute new points C and D, each at a ditance
# Clen off the line. Note that since ABperp is
# a vector with unit eEuclidean norm, if I
# multiply it by Clen, then it has length Clen.
C = ABmid + Clen*ABperp;
D = ABmid - Clen*ABperp;
# plot them all
plot(c(A[1],B[1]),c(A[2],B[2]),col='blue',type='l',asp=1)
lines(c(C[1],D[1]),c(C[2],D[2]),col='red')
# asp=1 ensures the lines appear mutually perpendicular. 
# If the axes had different units along the axes, then 
# the lines would look skewed.

Eugenio Alcala 2021-5-20

在 MATLAB Online 中打开

I think a more reliable answer to this problem is to use the rotation matrix concept.

So, assuming a set of points in a matrix where the first and second columns are X and Y, respectively, we can compute the angle of those points such as:

X1 = points(1,1);
X2 = points(2,1);
Y1 = points(1,2);
Y2 = points(2,2);
angle = atan2( (Y2-Y1), (X2-X1) );

From here we can compute the rotation matrix:

R = [cos(angle) -sin(angle);
     sin(angle) cos(angle) ];

And the new othogonal point that we are looking for at a distance "dist" to the left is:

XYnew = [X2; Y2] + R * [0; dist];

If we want another point to the right just recompute the line above with dist negative.

In a for loop it would look like the following:

for i=1:100
    X1 = points(i,1);
    X2 = points(i+1,1);
    Y1 = points(i,2);
    Y2 = points(i+1,2);
    
    angle(i) = atan2( (Y2-Y1), (X2-X1) );
    R = [cos(angle(i)) -sin(angle(i));
         sin(angle(i)) cos(angle(i))];
     
    XYnew = [X2; Y2] + R * [0;dist];
    x(i) = XYnew(1);
    y(i) = XYnew(2);
end