CODEWITHSUNDEEP
pythonpython-3.xlinedistancepoint
ROBlackSnail
ROBlackSnail

Reputation: 109

Distance from a point to a line

I have created a class "Point" and i want to calculate the shortest distance between a given point and a line ( characterized by 2 other points ), all points are known. I tried to use this formula : |Ax+By+C| / sqrt(A^2+B^2) , but i messed up and got more confused by the minute (mostly because of math formulas :( )...

I did find some sites where people asked this question too, but it either was not for Python or it was in a 3D system not 2D ...

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Below is my class : 

class Point:
        def __init__(self,initx,inity):
            self.x = initx
            self.y = inity
        def getX(self):
            return self.x
        def getY(self):
            return self.y
        def __str__(self):
            return "x=" + str(self.x) + ", y=" + str(self.y)
        def distance_from_point(self,the_other_point):
            dx = the_other_point.getX() - self.x
            dy = the_other_point.getY() - self.y
        def slope(self,other_point):
            if self.x - other_point.getX() == 0 :
                return 0
            else:
                panta = (self.y - other_point.getY())/ (self.x - other_point.getX())
                return panta

Can someone help me write a separate function or a method that does what i want ? I tried for 2 hours and I can't figure it out ...

Upvotes: 1

Views: 15478

Answers (4)

gboffi
gboffi

Reputation: 25023

enter image description here

It is

Q-P0 = α (P1-P0) + β k×(P1-P0)

where k×(P1-P0) is the vector multiplication between the z versor and the position vector.

Writing the above equation as a scalar system, we have

ΔX = α dx - β dy
ΔY = α dy + β dx

solving for α, β

(dx²+dy²) α = + ΔX dx + ΔY dy
(dx²+dy²) β = - ΔX dy + ΔY dx

Because (dx²+dy²) = |P1-P0|²

                        ΔY dx - ΔX dy
β |P1-P0| = distance = ---------------
                            |P1-P0|

Of course, our result is equivalent to (P1-P0)×(Q-P0)/|P1-P0|.

Final remark, the distance of Q from (P1-P0) is oriented, and maybe you need the absolute value of the distance.

Upvotes: 0

M.Ali Zarrinzade
M.Ali Zarrinzade

Reputation: 1

You can install FastLine via pip and then use it in this way:

from FastLine import Line
# define a line by two points
l1 = Line(p1=(0,0), p2=(10,10))
# or define a line by slope and intercept
l2 = Line(m=0.5, b=-1)

# compute distance
d1 = l1.distance_to((20,50))
# returns 21.213203435596427
d2 = l2.distance_to((-15,17))
# returns 22.807893370497855

Upvotes: 0

Iluvatar
Iluvatar

Reputation: 1537

You should be able to use this formula from the points directly. So, you'd have something like:

import math

class Point:
    def distance_to_line(self, p1, p2):
        x_diff = p2.x - p1.x
        y_diff = p2.y - p1.y
        num = abs(y_diff*self.x - x_diff*self.y + p2.x*p1.y - p2.y*p1.x)
        den = math.sqrt(y_diff**2 + x_diff**2)
        return num / den

Upvotes: 6

AnkurJat
AnkurJat

Reputation: 414

The distance formula between two points is Distance =sqrt((x2−x1)^2+(y2−y1)^2). And the formula to calculate slope is slope = (y2 - y1) / (x2 - x1).

so below is a simple method to calculate the distance

def distance_from_other_point(self, other_point):
    return math.sqrt( ( other_point.getX() - self.getX() )**2 + ( other_point.getY() - self.getY() )**2 )

def slope(self, otehr_point):
   return ( other_point.getY() - self.getY() )*1.0 / ( other_point.getX() - self.getX() )

In the second method, slope, I multiplied with 1.0 so that result will be in float. Note - I used the syntax of python 2.7.6 though hopefully, it will work in python 3.x as well.

Upvotes: 0

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