finding nearest perpendicular point to a line

Question

i am trying to find the perpendicular closest point to a line, given:

line is given by p1 and p2:
p1= [833974.3939975424, 10.741845010809124]
p2= [833972.1988621169, 9.989807874318076]

point: [833972.4500000001, 10.200000000000001]

I used these functions:


point1, point2 = nearest_points(shp.LineString([shp.Point(p1[0], p1[1]), shp.Point(p2[0], p2[1])]), shp.Point([point[0], point[1]]))

# and
line= shp.LineString([shp.Point(p1[0], p1[1]), shp.Point(p2[0], p2[1])])
pdist= line.project(shp.Point([point[0], point[1]]))
newpoint= line.interpolate(pdist)

when I use shapely functions, scikit-spatial, etc. I get the nearest non-perpendicular point as given in red (projected from green):

PLease, if anyone can help: I have been trying for hours to project the red point perpendicular to the line. I couldnt find a proper similar stack overflow answer.

Please provide your code and expected vs. actual outcome. Note the scale of your two axes are different, so it won't look perpendicular. — Mark Tolonen
– Mark Tolonen, Commented Jun 10 at 15:16
better say the line is defined by the blue (p1) and yellow (p2) points, and the additional point is the green one. As required above give your code — bruno
– bruno, Commented Jun 10 at 15:27
If q is the point off the line and p1, p2 the two points on the line then just project q-p1 onto p2-p1 with a scalar product. The resulting point is p1+[q-p1,p2-p1](p2-p1)/[p2-p1,p2-p1], where [,] is about the only inline way of writing a scalar product. — lastchance
– lastchance, Commented Jun 10 at 15:39

trincot · Accepted Answer · 2025-06-10 16:42:31Z

3

I stretched your image horizontally so that the visual distance of 0.5 units is the same horizontally as vertically:

Your calculation of the red point's coordinates seems to be fine.

answered Jun 10 at 16:42

trincot

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lastchance · Accepted Answer · 2025-06-10 15:53:22Z

2

The distance from p1 to the base of the perpendicular will just be the dot product of q-p1 with a unit direction vector along the displacement p2-p1. Then just multiply this by the unit vector and add to p1.

import numpy as np

def perpendicular( p1, p2, q ):
   return p1 + np.dot( q - p1, p2 - p1 ) * ( p2 - p1 ) / np.dot( p2 - p1, p2 - p1 )


p1 = np.array( [ 833974.3939975424, 10.741845010809124 ] )
p2 = np.array( [ 833972.1988621169,  9.989807874318076 ] )
q  = np.array( [ 833972.4500000001, 10.200000000000001 ] )

print( perpendicular( p1, p2, q ) )

Output:

[8.33972488e+05 1.00888871e+01]

As already pointed out in comments, this won't look right unless your axis scales are the same, so preserving orthogonality.

answered Jun 10 at 15:53

lastchance

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1 Comment

Sjoerd Braaksma Jun 10 at 16:05

ah thanks, so it's my axis scales that mess up so it doesn't look right!

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