## Check if a set of coordinates are on a line

### Sebastian Ursarescu (view profile)

on 18 Sep 2019
Latest activity Commented on by Sebastian Ursarescu

on 18 Sep 2019

### Bruno Luong (view profile)

I have a set of coordinates, for example x = [0 0 0 0] y = [1 0.9 1.1 1] or x = [1 2 3 4] y = [1 4 6 7], and I want to know if they are almost on a line. I can say the first exemple does and the second doesn't.
I tried with polyarea checking if the area is small enough but it's hard to define the crossing area value regarding the size of the polygon and, moreover, it gives the wrong answer if the polygon crosses itself, like in the first example.

KALYAN ACHARJYA

### KALYAN ACHARJYA (view profile)

on 18 Sep 2019
@Guillaume sir, sorry, I missed it, I misundestood the question and considering on image (Horizantal or Vertical line)
Always Thanks

### Bruno Luong (view profile)

on 18 Sep 2019
Edited by Bruno Luong

### Bruno Luong (view profile)

on 18 Sep 2019

xy=[x(:)';y(:)'];
criteriatol = 0.01; % adjust to your need, smaller means stricter line test
s = svd(xy-mean(xy,2));
isline = max(s)*criteriatol>min(s)
Work isotropically (orientation independent, but also scaling, shifting independent), even with vertical line.

#### 1 Comment

Sebastian Ursarescu

### Sebastian Ursarescu (view profile)

on 18 Sep 2019
Seems to work pretty good

### Guillaume (view profile)

on 18 Sep 2019

Probably, the best way is to use polyfit (which does a least square fit) and look at the norm of the residuals. If it's small enough your points (tolerance is up to you) are on a line.
[~, pfit] = polyfit(x, y, 1); %least square fit of a line
if pfit.normr <= arbitrary_value
disp('is a line');
else
disp('not a line');
end
You may need to special case when all the x are (almost) all equal (vertical line) if that's an actual possibility.

Sebastian Ursarescu

### Sebastian Ursarescu (view profile)

on 18 Sep 2019
The problem is actually when I have a vertical line. I already tried using functions but they don't cover the case of vertical line
Guillaume

### Guillaume (view profile)

on 18 Sep 2019
Then use Bruno's answer which is a lot better than mine anyway.