Finding the closest point on a line to another point

Hello all,

I am looking find the closest point on the line in the below image to each of the points shown by a circle.

I have 2 tables, one I will call line, and one I will call point:

line.x = [1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5]
line.y = [1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5]
point.x = [4.5, 2.1, 3.6]
point.y = [3.9, 2.7, 3.1]
scatter(point.x, point.y)
hold on
plot(line.x, line.y)

I want to find the closest defined point on the line (it has to be one of the points listed above, it can't be a point on the line between the defined points) that is closest to each of the circles.

Any tips on how to do this? I have found several similar threads but nothing that gets at what I'm after. Thanks all!!

[Turlough Hughes]

Are you treating the line as continuous or as the discrete points provided?

[Rae Taylor-Burns]

As the discrete points provided. Thanks!

Answer 1

[Turlough Hughes]

Solving for the discrete points is actually a lot easier. Using your data as follows:

line.x = [1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5];
line.y = [1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5];
point.x = [4.5, 2.1, 3.6];
point.y = [3.9, 2.7, 3.1];
h = figure(); plot(line.x,line.y,'ok','MarkerFaceColor','k');
hold on, plot(point.x,point.y,'or')
h.Position(3) = h.Position(4); % sets the figure aspect ratio to 1.

I just find the distance to each point in your line. The first row of dist is the distance of every point in line to the [point.x(1) point.y(1)], row two is all the distances to the second point and so on:

dist = sqrt((line.y - point.y.').^2 + (line.x - point.x.').^2)
[~,idx] = min(dist,[],2); % index for the closest points, i.e. those with min distance.

Plot resulting points on the line that are closest.

hold on; plot(line.x(idx),line.y(idx),'.cyan','MarkerSize',18)

[Rae Taylor-Burns]

Thanks! I think the problem with that is that the matrix dimensions are different - I get

Error using - . Matrix dimensions must agree.

Did you get the code above to work? I copied what you wrote and it did not work. Thank you for your help!

[Turlough Hughes]

Have you tried clearing your workspace and then entering the code exactly as provided above?

[James Tursa]

You can drop the sqrt( ) since it won't affect the resulting min( ) answer.

Answer 2

[Rae Taylor-Burns]

Ah yes - thank you!! Very helpful and fast!