Minimum distance from a point to the line segment using Vectors

Given the coordinates of two endpoints A(x1, y1), B(x2, y2) of the line segment and coordinates of a point E(x, y); the task is to find the minimum distance from the point to line segment formed with the given coordinates.
Note that both the ends of a line can go to infinity i.e. a line has no ending points. On the other hand, a line segment has start and endpoints due to which length of the line segment is fixed.
Examples:

Input: A = {0, 0}, B = {2, 0}, E = {4, 0} 

Output: 2 
To find the distance, dot product has to be found between vectors AB, BE and AB, AE. 
AB = (x2 – x1, y2 – y1) = (2 – 0, 0 – 0) = (2, 0) 
BE = (x – x2, y – y2) = (4 – 2, 0 – 0) = (2, 0) 
AE = (x – x1, y – y1) = (4 – 0, 0 – 0) = (4, 0) 
AB . BE = (AB_x * BE_x + AB_y * BE_y) = (2 * 2 + 0 * 0) = 4 
AB . AE = (AB_x * AE_x + AB_y * AE_y) = (2 * 4 + 0 * 0) = 8 
Therefore, nearest point from E to line segment is point B. 
Minimum Distance = BE =  = 2
Input: A = {0, 0}, B = {2, 0}, E = {1, 1} 
Output: 1 

Approach: The idea is to use the concept of vectors to solve the problem since the nearest point always lies on the line segment. Assuming that the direction of vector AB is A to B, there are three cases that arise:  

1. The nearest point from the point E on the line segment AB is point B itself if the dot product of vector AB(A to B) and vector BE(B to E) is positive where E is the given point. Since AB . BE > 0, the given point lies in the same direction as the vector AB is and the nearest point must be B itself because the nearest point lies on the line segment. 

2. The nearest point from the point E on the line segment AB is point A itself if the dot product of vector AB(A to B) and vector AE(A to E) is negative where E is the given point. Since AB . AE < 0, the given point lies in the opposite direction of the line segment AB and the nearest point must be A itself because the nearest point lies on the line segment. 

3. Otherwise, if the two dot products AB.BE and AB.AE are of different signs, then the point E is perpendicular to the line segment AB and the perpendicular distance to the given point E from the line segment AB is the shortest distance. If some arbitrary point F is the point on the line segment which is perpendicular to E, then the perpendicular distance can be calculated as |EF| = |(AB X AE)/|AB||

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>

// To store the point
#define Point pair<double, double>
#define F first
#define S second
using namespace std;

// Function to return the minimum distance
// between a line segment AB and a point E
double minDistance(Point A, Point B, Point E)
{

    // vector AB
    pair<double, double> AB;
    AB.F = B.F - A.F;
    AB.S = B.S - A.S;

    // vector BP
    pair<double, double> BE;
    BE.F = E.F - B.F;
    BE.S = E.S - B.S;

    // vector AP
    pair<double, double> AE;
    AE.F = E.F - A.F,
    AE.S = E.S - A.S;

    // Variables to store dot product
    double AB_BE, AB_AE;

    // Calculating the dot product
    AB_BE = (AB.F * BE.F + AB.S * BE.S);
    AB_AE = (AB.F * AE.F + AB.S * AE.S);

    // Minimum distance from
    // point E to the line segment
    double reqAns = 0;

    // Case 1
    if (AB_BE > 0) {

        // Finding the magnitude
        double y = E.S - B.S;
        double x = E.F - B.F;
        reqAns = sqrt(x * x + y * y);
    }

    // Case 2
    else if (AB_AE < 0) {
        double y = E.S - A.S;
        double x = E.F - A.F;
        reqAns = sqrt(x * x + y * y);
    }

    // Case 3
    else {

        // Finding the perpendicular distance
        double x1 = AB.F;
        double y1 = AB.S;
        double x2 = AE.F;
        double y2 = AE.S;
        double mod = sqrt(x1 * x1 + y1 * y1);
        reqAns = abs(x1 * y2 - y1 * x2) / mod;
    }
    return reqAns;
}

// Driver code
int main()
{
    Point A = make_pair(0, 0);
    Point B = make_pair(2, 0);
    Point E = make_pair(1, 1);

    cout << minDistance(A, B, E);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
class GFG
{

static class pair
{
    double F, S;
    public pair(double F, double S)
    {
        this.F = F;
        this.S = S;
    }
    public pair() {
    }
}

// Function to return the minimum distance
// between a line segment AB and a point E
static double minDistance(pair A, pair B, pair E)
{

    // vector AB
    pair AB = new pair();
    AB.F = B.F - A.F;
    AB.S = B.S - A.S;

    // vector BP
    pair BE = new pair();
    BE.F = E.F - B.F;
    BE.S = E.S - B.S;

    // vector AP
    pair AE = new pair();
    AE.F = E.F - A.F;
    AE.S = E.S - A.S;

    // Variables to store dot product
    double AB_BE, AB_AE;

    // Calculating the dot product
    AB_BE = (AB.F * BE.F + AB.S * BE.S);
    AB_AE = (AB.F * AE.F + AB.S * AE.S);

    // Minimum distance from
    // point E to the line segment
    double reqAns = 0;

    // Case 1
    if (AB_BE > 0)
    {

        // Finding the magnitude
        double y = E.S - B.S;
        double x = E.F - B.F;
        reqAns = Math.sqrt(x * x + y * y);
    }

    // Case 2
    else if (AB_AE < 0)
    {
        double y = E.S - A.S;
        double x = E.F - A.F;
        reqAns = Math.sqrt(x * x + y * y);
    }

    // Case 3
    else
    {

        // Finding the perpendicular distance
        double x1 = AB.F;
        double y1 = AB.S;
        double x2 = AE.F;
        double y2 = AE.S;
        double mod = Math.sqrt(x1 * x1 + y1 * y1);
        reqAns = Math.abs(x1 * y2 - y1 * x2) / mod;
    }
    return reqAns;
}

// Driver code
public static void main(String[] args)
{
    pair A = new pair(0, 0);
    pair B = new pair(2, 0);
    pair E = new pair(1, 1);

    System.out.print((int)minDistance(A, B, E));
}
}

// This code is contributed by 29AjayKumar

C#

// C# implementation of the approach
using System;

class GFG
{

class pair
{
    public double F, S;
    public pair(double F, double S)
    {
        this.F = F;
        this.S = S;
    }
    public pair() {
    }
}

// Function to return the minimum distance
// between a line segment AB and a point E
static double minDistance(pair A, pair B, pair E)
{

    // vector AB
    pair AB = new pair();
    AB.F = B.F - A.F;
    AB.S = B.S - A.S;

    // vector BP
    pair BE = new pair();
    BE.F = E.F - B.F;
    BE.S = E.S - B.S;

    // vector AP
    pair AE = new pair();
    AE.F = E.F - A.F;
    AE.S = E.S - A.S;

    // Variables to store dot product
    double AB_BE, AB_AE;

    // Calculating the dot product
    AB_BE = (AB.F * BE.F + AB.S * BE.S);
    AB_AE = (AB.F * AE.F + AB.S * AE.S);

    // Minimum distance from
    // point E to the line segment
    double reqAns = 0;

    // Case 1
    if (AB_BE > 0)
    {

        // Finding the magnitude
        double y = E.S - B.S;
        double x = E.F - B.F;
        reqAns = Math.Sqrt(x * x + y * y);
    }

    // Case 2
    else if (AB_AE < 0)
    {
        double y = E.S - A.S;
        double x = E.F - A.F;
        reqAns = Math.Sqrt(x * x + y * y);
    }

    // Case 3
    else
    {

        // Finding the perpendicular distance
        double x1 = AB.F;
        double y1 = AB.S;
        double x2 = AE.F;
        double y2 = AE.S;
        double mod = Math.Sqrt(x1 * x1 + y1 * y1);
        reqAns = Math.Abs(x1 * y2 - y1 * x2) / mod;
    }
    return reqAns;
}

// Driver code
public static void Main(String[] args)
{
    pair A = new pair(0, 0);
    pair B = new pair(2, 0);
    pair E = new pair(1, 1);

    Console.Write((int)minDistance(A, B, E));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript implementation of the approach

// Function to return the minimum distance
// between a line segment AB and a point E
function minDistance( A,  B,  E)
{

    // vector AB
    var AB=[];
    AB.push (B[0] - A[0]);
    AB.push(B[1] - A[1]);

    // vector BP
    var BE=[];
    BE.push(E[0] - B[0]);
    BE.push(E[1] - B[1]);

    // vector AP
   var AE=[];
    AE.push(E[0] - A[0]),
    AE.push(E[1] - A[1]);

    // Variables to store dot product
    var AB_BE, AB_AE;

    // Calculating the dot product
    AB_BE=(AB[0] * BE[0] + AB[1] * BE[1]);
    AB_AE=(AB[0] * AE[0] + AB[1] * AE[1]);

    // Minimum distance from
    // point E to the line segment
    var reqAns = 0;

    // Case 1
    if (AB_BE > 0) {

        // Finding the magnitude
        var y = E[1] - B[1];
        var x = E[0] - B[0];
        reqAns = Math.sqrt(x * x + y * y);
    }

    // Case 2
    else if (AB_AE < 0) {
        var y = E[1] - A[1];
        var x = E[0] - A[0];
        reqAns = Math.sqrt(x * x + y * y);
    }

    // Case 3
    else {

        // Finding the perpendicular distance
       var x1 = AB[0];
        var y1 = AB[1];
       var x2 = AE[0];
        var y2 = AE[1];
        var mod = Math.sqrt(x1 * x1 + y1 * y1);
        reqAns = Math.abs(x1 * y2 - y1 * x2) / mod;
    }
    return reqAns;
}

var A =[0, 0];
    var B = [2, 0];
    var E = [1, 1];

    document.write( minDistance(A, B, E));


</script>

Python3

# Python3 implementation of the approach
from math import sqrt

# Function to return the minimum distance
# between a line segment AB and a point E
def minDistance(A, B, E) :

    # vector AB
    AB = [None, None];
    AB[0] = B[0] - A[0];
    AB[1] = B[1] - A[1];

    # vector BP
    BE = [None, None];
    BE[0] = E[0] - B[0];
    BE[1] = E[1] - B[1];

    # vector AP
    AE = [None, None];
    AE[0] = E[0] - A[0];
    AE[1] = E[1] - A[1];

    # Variables to store dot product

    # Calculating the dot product
    AB_BE = AB[0] * BE[0] + AB[1] * BE[1];
    AB_AE = AB[0] * AE[0] + AB[1] * AE[1];

    # Minimum distance from
    # point E to the line segment
    reqAns = 0;

    # Case 1
    if (AB_BE > 0) :

        # Finding the magnitude
        y = E[1] - B[1];
        x = E[0] - B[0];
        reqAns = sqrt(x * x + y * y);

    # Case 2
    elif (AB_AE < 0) :
        y = E[1] - A[1];
        x = E[0] - A[0];
        reqAns = sqrt(x * x + y * y);

    # Case 3
    else:

        # Finding the perpendicular distance
        x1 = AB[0];
        y1 = AB[1];
        x2 = AE[0];
        y2 = AE[1];
        mod = sqrt(x1 * x1 + y1 * y1);
        reqAns = abs(x1 * y2 - y1 * x2) / mod;

    return reqAns;

# Driver code
if __name__ == "__main__" :

    A = [0, 0];
    B = [2, 0];
    E = [1, 1];

    print(minDistance(A, B, E));

# This code is contributed by AnkitRai01

Output

1

Time Complexity: O(log(x²+y²)) because it is using inbuilt sqrt function
Auxiliary Space: O(1)