how to find the distance from a point to a line

I am ask to find the distance from a point to a line

with this given

(0,0,12); x=4t, y=-2t, z=2t

but I don't know to to encode it to MATLAB

[Image Analyst]

I should have known. First time posters almost always delete their post.

[Rik]

I recovered the removed content from the Google cache (something which anyone can do). Editing away your question is very rude. Someone spent time reading your question, understanding your issue, figuring out the solution, and writing an answer. Now you repay that kindness by ensuring that the next person with a similar question can't benefit from this answer.

[Image Analyst]

Thanks Rik. Here is is again in case it gets deleted again:

I am ask to find the distance from a point to a line

with this given

(0,0,12); x=4t, y=-2t, z=2t

but I don't know to to encode it to MATLAB

I've tried recovering deleted stuff from archives but they either say they don't have that URL or else they have the latest one with deleted content.

[Rik]

No problem.

The Google search result layout has changed a bit in the interim, but I don't tend to have problems. I must admit I never look for a capture in the Wayback Machine though.

[Image Analyst]

I use a Firefox plug-in/add-on called WebArchives that searched Google plus several other archives. It's a icon on the Firefox toolbar (once installed).

[Rik]

I wonder why a normal search turns up a cached page, but this plug-in doesn't. Otherwise this would make it a lot easier.

(for the record: it is also available for Chromium-based browsers)

Answer 1

[Image Analyst]

See attached demo

% Get the distance from a point (x3, y3) to
% a line defined by two points (x1, y1) and (x2, y2);
% Reference: http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
function distance = GetPointLineDistance(x3,y3,x1,y1,x2,y2)
try
	
	% Find the numerator for our point-to-line distance formula.
	numerator = abs((x2 - x1) * (y1 - y3) - (x1 - x3) * (y2 - y1));
	
	% Find the denominator for our point-to-line distance formula.
	denominator = sqrt((x2 - x1) ^ 2 + (y2 - y1) ^ 2);
	
	% Compute the distance.
	distance = numerator ./ denominator;
catch ME
	callStackString = GetCallStack(ME);
	errorMessage = sprintf('Error in program %s.\nTraceback (most recent at top):\n%s\nError Message:\n%s',...
		mfilename, callStackString, ME.message);
	uiwait(warndlg(errorMessage))
end
return; % from GetPointLineDistance()
end

[Furkan Bek]

Nice implementation of the triangle theory.

This was proved to be working just fine. Thanks!

[Maite Osaba]

Would this work if the coordinates were expressed as longitude and latitude? Would it be enough to then convert the distance using the deg2km?

[Rik]

@Maite Osaba, latitude and longitude can be very non-linear. If you want accuracy, I would suggest not taking shortcuts.

[Image Analyst]

@Maite Osaba it is probably good for short distances where the earth is fairly flat. Of course it's the straight line distance and if the points are very widely spaced, you're getting the straight line distances, which would mean boring through the earth, if the earth is significantly curved over that distance. If you want to do this in general, then I'm sure there are probably functions in the Mapping Toolbox to help you though I'm not sure what they are since I don't have that toolbox. (Expand comments to see @Rik's comments just before mine.)