Related Articles
Minimum distance from a point to the line segment using Vectors
• Difficulty Level : Easy
• Last Updated : 10 Feb, 2021

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 end points due to which length of the line segment is fixed.
Examples:

Input: A = {0, 0}, B = {2, 0}, E = {4, 0}
Output:
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 = (ABx * BEx + ABy * BEy) = (2 * 2 + 0 * 0) = 4
AB . AE = (ABx * AEx + ABy * AEy) = (2 * 4 + 0 * 0) = 8
Therefore, nearest point from E to line segment is point B.
Minimum Distance = BE =

$\sqrt{(E_{y}&space;-&space;B_{y})^2&space;-&space;(E_x&space;-&space;B_x)^2}$

= 2
Input: A = {0, 0}, B = {2, 0}, E = {1, 1}
Output:

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 BE(B to E) is negative where E is the given point. Since AB . BE < 0, the given point lies on 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. If the dot product is 0, 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  // To store the point#define Point pair#define F first#define S secondusing namespace std; // Function to return the minimum distance// between a line segment AB and a point Edouble 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 codeint 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 approachclass 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 Estatic 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 codepublic 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

## Python3

 # Python3 implementation of the approachfrom math import sqrt # Function to return the minimum distance# between a line segment AB and a point Edef 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 codeif __name__ == "__main__" :     A = [0, 0];    B = [2, 0];    E = [1, 1];     print(minDistance(A, B, E)); # This code is contributed by AnkitRai01

## C#

 // C# implementation of the approachusing 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 Estatic 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 codepublic 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
Output:
1

Time Complexity: O(1 )

Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up