How to check if a given point lies inside or outside a polygon?

Given a polygon and a point ‘p‘, find if ‘p‘ lies inside the polygon or not. The points lying on the border are considered inside.

Examples:

Approach: The idea to solve this problem is based on How to check if two given line segments intersect, and to be used as follows:

1. Draw a horizontal line to the right of each point and extend it to infinity
2. Count the number of times the line intersects with polygon edges.
3. A point is inside the polygon if either count of intersections is odd or point lies on an edge of polygon.  If none of the conditions is true, then point lies outside.

How to handle point ‘g’ in the above figure?

Note that we should return true if the point lies on the line or same as one of the vertices of the given polygon. To handle this, after checking if the line from ‘p’ to extreme intersects, we check whether ‘p’ is collinear with vertices of current line of polygon. If it is collinear, then we check if the point ‘p’ lies on current side of polygon, if it lies, we return true, else false.

Following is the implementation of the above approach: 

C++

// A C++ program to check if a given point lies inside a given polygon
// Refer https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/
// for explanation of functions onSegment(), orientation() and doIntersect()
#include <iostream>
using namespace std;

// Define Infinite (Using INT_MAX caused overflow problems)
#define INF 10000

struct Point
{
    int x;
    int y;
};

// Given three collinear points p, q, r, the function checks if
// point q lies on line segment 'pr'
bool onSegment(Point p, Point q, Point r)
{
    if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) &&
            q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y))
        return true;
    return false;
}

// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are collinear
// 1 --> Clockwise
// 2 --> Counterclockwise
int orientation(Point p, Point q, Point r)
{
    int val = (q.y - p.y) * (r.x - q.x) -
            (q.x - p.x) * (r.y - q.y);

    if (val == 0) return 0; // collinear
    return (val > 0)? 1: 2; // clock or counterclock wise
}

// The function that returns true if line segment 'p1q1'
// and 'p2q2' intersect.
bool doIntersect(Point p1, Point q1, Point p2, Point q2)
{
    // Find the four orientations needed for general and
    // special cases
    int o1 = orientation(p1, q1, p2);
    int o2 = orientation(p1, q1, q2);
    int o3 = orientation(p2, q2, p1);
    int o4 = orientation(p2, q2, q1);

    // General case
    if (o1 != o2 && o3 != o4)
        return true;

    // Special Cases
    // p1, q1 and p2 are collinear and p2 lies on segment p1q1
    if (o1 == 0 && onSegment(p1, p2, q1)) return true;

    // p1, q1 and p2 are collinear and q2 lies on segment p1q1
    if (o2 == 0 && onSegment(p1, q2, q1)) return true;

    // p2, q2 and p1 are collinear and p1 lies on segment p2q2
    if (o3 == 0 && onSegment(p2, p1, q2)) return true;

    // p2, q2 and q1 are collinear and q1 lies on segment p2q2
    if (o4 == 0 && onSegment(p2, q1, q2)) return true;

    return false; // Doesn't fall in any of the above cases
}

// Returns true if the point p lies inside the polygon[] with n vertices
bool isInside(Point polygon[], int n, Point p)
{
    // There must be at least 3 vertices in polygon[]
    if (n < 3) return false;

    // Create a point for line segment from p to infinite
    Point extreme = {INF, p.y};

      // To count number of points in polygon
      // whose y-coordinate is equal to
      // y-coordinate of the point
      int decrease = 0;

    // Count intersections of the above line with sides of polygon
    int count = 0, i = 0;
    do
    {
        int next = (i+1)%n;

          if(polygon[i].y == p.y) decrease++;

        // Check if the line segment from 'p' to 'extreme' intersects
        // with the line segment from 'polygon[i]' to 'polygon[next]'
        if (doIntersect(polygon[i], polygon[next], p, extreme))
        {
            // If the point 'p' is collinear with line segment 'i-next',
            // then check if it lies on segment. If it lies, return true,
            // otherwise false
            if (orientation(polygon[i], p, polygon[next]) == 0)
            return onSegment(polygon[i], p, polygon[next]);

            count++;
        }
        i = next;
    } while (i != 0);

  // Reduce the count by decrease amount
  // as these points would have been added twice
      count -= decrease;

    // Return true if count is odd, false otherwise
    return count&1; // Same as (count%2 == 1)
}

// Driver program to test above functions
int main()
{
    Point polygon1[] = {{0, 0}, {10, 0}, {10, 10}, {0, 10}};
    int n = sizeof(polygon1)/sizeof(polygon1[0]);
    Point p = {20, 20};
    isInside(polygon1, n, p)? cout << "Yes \n": cout << "No \n";

    p = {5, 5};
    isInside(polygon1, n, p)? cout << "Yes \n": cout << "No \n";

    Point polygon2[] = {{0, 0}, {5, 5}, {5, 0}};
    p = {3, 3};
    n = sizeof(polygon2)/sizeof(polygon2[0]);
    isInside(polygon2, n, p)? cout << "Yes \n": cout << "No \n";

    p = {5, 1};
    isInside(polygon2, n, p)? cout << "Yes \n": cout << "No \n";

    p = {8, 1};
    isInside(polygon2, n, p)? cout << "Yes \n": cout << "No \n";

    Point polygon3[] = {{0, 0}, {10, 0}, {10, 10}, {0, 10}};
    p = {-1,10};
    n = sizeof(polygon3)/sizeof(polygon3[0]);
    isInside(polygon3, n, p)? cout << "Yes \n": cout << "No \n";

    return 0;
}

Java

// A Java program to check if a given point
// lies inside a given polygon
// Refer https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/
// for explanation of functions onSegment(),
// orientation() and doIntersect()
class GFG
{

    // Define Infinite (Using INT_MAX
    // caused overflow problems)
    static int INF = 10000;

    static class Point
    {
        int x;
        int y;

        public Point(int x, int y)
        {
            this.x = x;
            this.y = y;
        }
    };

    // Given three collinear points p, q, r,
    // the function checks if point q lies
    // on line segment 'pr'
    static boolean onSegment(Point p, Point q, Point r)
    {
        if (q.x <= Math.max(p.x, r.x) &&
            q.x >= Math.min(p.x, r.x) &&
            q.y <= Math.max(p.y, r.y) &&
            q.y >= Math.min(p.y, r.y))
        {
            return true;
        }
        return false;
    }

    // To find orientation of ordered triplet (p, q, r).
    // The function returns following values
    // 0 --> p, q and r are collinear
    // 1 --> Clockwise
    // 2 --> Counterclockwise
    static int orientation(Point p, Point q, Point r)
    {
        int val = (q.y - p.y) * (r.x - q.x)
                - (q.x - p.x) * (r.y - q.y);

        if (val == 0)
        {
            return 0; // collinear
        }
        return (val > 0) ? 1 : 2; // clock or counterclock wise
    }

    // The function that returns true if
    // line segment 'p1q1' and 'p2q2' intersect.
    static boolean doIntersect(Point p1, Point q1,
                            Point p2, Point q2)
    {
        // Find the four orientations needed for
        // general and special cases
        int o1 = orientation(p1, q1, p2);
        int o2 = orientation(p1, q1, q2);
        int o3 = orientation(p2, q2, p1);
        int o4 = orientation(p2, q2, q1);

        // General case
        if (o1 != o2 && o3 != o4)
        {
            return true;
        }

        // Special Cases
        // p1, q1 and p2 are collinear and
        // p2 lies on segment p1q1
        if (o1 == 0 && onSegment(p1, p2, q1))
        {
            return true;
        }

        // p1, q1 and p2 are collinear and
        // q2 lies on segment p1q1
        if (o2 == 0 && onSegment(p1, q2, q1))
        {
            return true;
        }

        // p2, q2 and p1 are collinear and
        // p1 lies on segment p2q2
        if (o3 == 0 && onSegment(p2, p1, q2))
        {
            return true;
        }

        // p2, q2 and q1 are collinear and
        // q1 lies on segment p2q2
        if (o4 == 0 && onSegment(p2, q1, q2))
        {
            return true;
        }

        // Doesn't fall in any of the above cases
        return false;
    }

    // Returns true if the point p lies
    // inside the polygon[] with n vertices
    static boolean isInside(Point polygon[], int n, Point p)
    {
        // There must be at least 3 vertices in polygon[]
        if (n < 3)
        {
            return false;
        }

        // Create a point for line segment from p to infinite
        Point extreme = new Point(INF, p.y);

          // To count number of points in polygon
          // whose y-coordinate is equal to
          // y-coordinate of the point
          int decrease = 0;

        // Count intersections of the above line
        // with sides of polygon
        int count = 0, i = 0;
        do
        {
            int next = (i + 1) % n;

              if(polygon[i].y == p.y) decrease++;

            // Check if the line segment from 'p' to
            // 'extreme' intersects with the line
            // segment from 'polygon[i]' to 'polygon[next]'
            if (doIntersect(polygon[i], polygon[next], p, extreme))
            {
                // If the point 'p' is collinear with line
                // segment 'i-next', then check if it lies
                // on segment. If it lies, return true, otherwise false
                if (orientation(polygon[i], p, polygon[next]) == 0)
                {
                    return onSegment(polygon[i], p,
                                    polygon[next]);
                }

                count++;
            }
            i = next;
        } while (i != 0);

          // Reduce the count by decrease amount
          // as these points would have been added twice
          count -= decrease;

        // Return true if count is odd, false otherwise
        return (count % 2 == 1); // Same as (count%2 == 1)
    }

    // Driver Code
    public static void main(String[] args)
    {
        Point polygon1[] = {new Point(0, 0),
                            new Point(10, 0),
                            new Point(10, 10),
                            new Point(0, 10)};
        int n = polygon1.length;
        Point p = new Point(20, 20);
        if (isInside(polygon1, n, p))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
        p = new Point(5, 5);
        if (isInside(polygon1, n, p))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
        Point polygon2[] = {new Point(0, 0),
            new Point(5, 5), new Point(5, 0)};
        p = new Point(3, 3);
        n = polygon2.length;
        if (isInside(polygon2, n, p))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
        p = new Point(5, 1);
        if (isInside(polygon2, n, p))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
        p = new Point(8, 1);
        if (isInside(polygon2, n, p))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
        Point polygon3[] = {new Point(0, 0),
                            new Point(10, 0),
                            new Point(10, 10),
                            new Point(0, 10)};
        p = new Point(-1, 10);
        n = polygon3.length;
        if (isInside(polygon3, n, p))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
    }
}

// This code is contributed by 29AjayKumar

Python3

# A Python3 program to check if a given point
# lies inside a given polygon
# Refer https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/
# for explanation of functions onSegment(),
# orientation() and doIntersect()

# Define Infinite (Using INT_MAX
# caused overflow problems)
INT_MAX = 10000

# Given three collinear points p, q, r,
# the function checks if point q lies
# on line segment 'pr'
def onSegment(p:tuple, q:tuple, r:tuple) -> bool:

    if ((q[0] <= max(p[0], r[0])) &
        (q[0] >= min(p[0], r[0])) &
        (q[1] <= max(p[1], r[1])) &
        (q[1] >= min(p[1], r[1]))):
        return True

    return False

# To find orientation of ordered triplet (p, q, r).
# The function returns following values
# 0 --> p, q and r are collinear
# 1 --> Clockwise
# 2 --> Counterclockwise
def orientation(p:tuple, q:tuple, r:tuple) -> int:

    val = (((q[1] - p[1]) *
            (r[0] - q[0])) -
           ((q[0] - p[0]) *
            (r[1] - q[1])))

    if val == 0:
        return 0
    if val > 0:
        return 1 # Collinear
    else:
        return 2 # Clock or counterclock

def doIntersect(p1, q1, p2, q2):

    # Find the four orientations needed for
    # general and special cases
    o1 = orientation(p1, q1, p2)
    o2 = orientation(p1, q1, q2)
    o3 = orientation(p2, q2, p1)
    o4 = orientation(p2, q2, q1)

    # General case
    if (o1 != o2) and (o3 != o4):
        return True

    # Special Cases
    # p1, q1 and p2 are collinear and
    # p2 lies on segment p1q1
    if (o1 == 0) and (onSegment(p1, p2, q1)):
        return True

    # p1, q1 and p2 are collinear and
    # q2 lies on segment p1q1
    if (o2 == 0) and (onSegment(p1, q2, q1)):
        return True

    # p2, q2 and p1 are collinear and
    # p1 lies on segment p2q2
    if (o3 == 0) and (onSegment(p2, p1, q2)):
        return True

    # p2, q2 and q1 are collinear and
    # q1 lies on segment p2q2
    if (o4 == 0) and (onSegment(p2, q1, q2)):
        return True

    return False

# Returns true if the point p lies
# inside the polygon[] with n vertices
def is_inside_polygon(points:list, p:tuple) -> bool:

    n = len(points)

    # There must be at least 3 vertices
    # in polygon
    if n < 3:
        return False

    # Create a point for line segment
    # from p to infinite
    extreme = (INT_MAX, p[1])

    # To count number of points in polygon
      # whose y-coordinate is equal to
      # y-coordinate of the point
    decrease = 0
    count = i = 0

    while True:
        next = (i + 1) % n

        if(points[i][1] == p[1]):
            decrease += 1

        # Check if the line segment from 'p' to
        # 'extreme' intersects with the line
        # segment from 'polygon[i]' to 'polygon[next]'
        if (doIntersect(points[i],
                        points[next],
                        p, extreme)):

            # If the point 'p' is collinear with line
            # segment 'i-next', then check if it lies
            # on segment. If it lies, return true, otherwise false
            if orientation(points[i], p,
                           points[next]) == 0:
                return onSegment(points[i], p,
                                 points[next])

            count += 1

        i = next

        if (i == 0):
            break

    # Reduce the count by decrease amount
      # as these points would have been added twice
    count -= decrease

    # Return true if count is odd, false otherwise
    return (count % 2 == 1)

# Driver code
if __name__ == '__main__':

    polygon1 = [ (0, 0), (10, 0), (10, 10), (0, 10) ]

    p = (20, 20)
    if (is_inside_polygon(points = polygon1, p = p)):
      print ('Yes')
    else:
      print ('No')

    p = (5, 5)
    if (is_inside_polygon(points = polygon1, p = p)):
      print ('Yes')
    else:
      print ('No')

    polygon2 = [ (0, 0), (5, 0), (5, 5), (3, 3) ]

    p = (3, 3)
    if (is_inside_polygon(points = polygon2, p = p)):
      print ('Yes')
    else:
      print ('No')

    p = (5, 1)
    if (is_inside_polygon(points = polygon2, p = p)):
      print ('Yes')
    else:
      print ('No')

    p = (8, 1)
    if (is_inside_polygon(points = polygon2, p = p)):
      print ('Yes')
    else:
      print ('No')

    polygon3 = [ (0, 0), (10, 0), (10, 10), (0, 10) ]

    p = (-1, 10)
    if (is_inside_polygon(points = polygon3, p = p)):
      print ('Yes')
    else:
      print ('No')

# This code is contributed by Vikas Chitturi

C#

// A C# program to check if a given point
// lies inside a given polygon
// Refer https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/
// for explanation of functions onSegment(),
// orientation() and doIntersect()
using System;

class GFG
{

    // Define Infinite (Using INT_MAX
    // caused overflow problems)
    static int INF = 10000;

    class Point
    {
        public int x;
        public int y;

        public Point(int x, int y)
        {
            this.x = x;
            this.y = y;
        }
    };

    // Given three collinear points p, q, r,
    // the function checks if point q lies
    // on line segment 'pr'
    static bool onSegment(Point p, Point q, Point r)
    {
        if (q.x <= Math.Max(p.x, r.x) &&
            q.x >= Math.Min(p.x, r.x) &&
            q.y <= Math.Max(p.y, r.y) &&
            q.y >= Math.Min(p.y, r.y))
        {
            return true;
        }
        return false;
    }

    // To find orientation of ordered triplet (p, q, r).
    // The function returns following values
    // 0 --> p, q and r are collinear
    // 1 --> Clockwise
    // 2 --> Counterclockwise
    static int orientation(Point p, Point q, Point r)
    {
        int val = (q.y - p.y) * (r.x - q.x) -
                (q.x - p.x) * (r.y - q.y);

        if (val == 0)
        {
            return 0; // collinear
        }
        return (val > 0) ? 1 : 2; // clock or counterclock wise
    }

    // The function that returns true if
    // line segment 'p1q1' and 'p2q2' intersect.
    static bool doIntersect(Point p1, Point q1,
                            Point p2, Point q2)
    {
        // Find the four orientations needed for
        // general and special cases
        int o1 = orientation(p1, q1, p2);
        int o2 = orientation(p1, q1, q2);
        int o3 = orientation(p2, q2, p1);
        int o4 = orientation(p2, q2, q1);

        // General case
        if (o1 != o2 && o3 != o4)
        {
            return true;
        }

        // Special Cases
        // p1, q1 and p2 are collinear and
        // p2 lies on segment p1q1
        if (o1 == 0 && onSegment(p1, p2, q1))
        {
            return true;
        }

        // p1, q1 and p2 are collinear and
        // q2 lies on segment p1q1
        if (o2 == 0 && onSegment(p1, q2, q1))
        {
            return true;
        }

        // p2, q2 and p1 are collinear and
        // p1 lies on segment p2q2
        if (o3 == 0 && onSegment(p2, p1, q2))
        {
            return true;
        }

        // p2, q2 and q1 are collinear and
        // q1 lies on segment p2q2
        if (o4 == 0 && onSegment(p2, q1, q2))
        {
            return true;
        }

        // Doesn't fall in any of the above cases
        return false;
    }

    // Returns true if the point p lies
    // inside the polygon[] with n vertices
    static bool isInside(Point []polygon, int n, Point p)
    {
        // There must be at least 3 vertices in polygon[]
        if (n < 3)
        {
            return false;
        }

        // Create a point for line segment from p to infinite
        Point extreme = new Point(INF, p.y);

        // Count intersections of the above line
        // with sides of polygon
        int count = 0, i = 0;
        do
        {
            int next = (i + 1) % n;

            // Check if the line segment from 'p' to
            // 'extreme' intersects with the line
            // segment from 'polygon[i]' to 'polygon[next]'
            if (doIntersect(polygon[i],
                            polygon[next], p, extreme))
            {
                // If the point 'p' is collinear with line
                // segment 'i-next', then check if it lies
                // on segment. If it lies, return true, otherwise false
                if (orientation(polygon[i], p, polygon[next]) == 0)
                {
                    return onSegment(polygon[i], p,
                                    polygon[next]);
                }
                count++;
            }
            i = next;
        } while (i != 0);

        // Return true if count is odd, false otherwise
        return (count % 2 == 1); // Same as (count%2 == 1)
    }

    // Driver Code
    public static void Main(String[] args)
    {
        Point []polygon1 = {new Point(0, 0),
                            new Point(10, 0),
                            new Point(10, 10),
                            new Point(0, 10)};
        int n = polygon1.Length;
        Point p = new Point(20, 20);
        if (isInside(polygon1, n, p))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
        p = new Point(5, 5);
        if (isInside(polygon1, n, p))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
        Point []polygon2 = {new Point(0, 0),
                            new Point(5, 5),
                            new Point(5, 0)};
        p = new Point(3, 3);
        n = polygon2.Length;
        if (isInside(polygon2, n, p))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
        p = new Point(5, 1);
        if (isInside(polygon2, n, p))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
        p = new Point(8, 1);
        if (isInside(polygon2, n, p))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
        Point []polygon3 = {new Point(0, 0),
                            new Point(10, 0),
                            new Point(10, 10),
                            new Point(0, 10)};
        p = new Point(-1, 10);
        n = polygon3.Length;
        if (isInside(polygon3, n, p))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// A Javascript program to check if a given point
// lies inside a given polygon
// Refer https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/
// for explanation of functions onSegment(),
// orientation() and doIntersect()

// Define Infinite (Using INT_MAX
    // caused overflow problems)
let INF = 10000;

class Point
{
    constructor(x,y)
    {
        this.x = x;
        this.y = y;
    }
}

// Given three collinear points p, q, r,
    // the function checks if point q lies
    // on line segment 'pr'
function onSegment(p,q,r)
{
     if (q.x <= Math.max(p.x, r.x) &&
            q.x >= Math.min(p.x, r.x) &&
            q.y <= Math.max(p.y, r.y) &&
            q.y >= Math.min(p.y, r.y))
        {
            return true;
        }
        return false;
}

// To find orientation of ordered triplet (p, q, r).
    // The function returns following values
    // 0 --> p, q and r are collinear
    // 1 --> Clockwise
    // 2 --> Counterclockwise
function orientation(p,q,r)
{
    let val = (q.y - p.y) * (r.x - q.x)
                - (q.x - p.x) * (r.y - q.y);

        if (val == 0)
        {
            return 0; // collinear
        }
        return (val > 0) ? 1 : 2; // clock or counterclock wise
}

// The function that returns true if
    // line segment 'p1q1' and 'p2q2' intersect.
function  doIntersect(p1,q1,p2,q2)
{
    // Find the four orientations needed for
        // general and special cases
        let o1 = orientation(p1, q1, p2);
        let o2 = orientation(p1, q1, q2);
        let o3 = orientation(p2, q2, p1);
        let o4 = orientation(p2, q2, q1);

        // General case
        if (o1 != o2 && o3 != o4)
        {
            return true;
        }

        // Special Cases
        // p1, q1 and p2 are collinear and
        // p2 lies on segment p1q1
        if (o1 == 0 && onSegment(p1, p2, q1))
        {
            return true;
        }

        // p1, q1 and p2 are collinear and
        // q2 lies on segment p1q1
        if (o2 == 0 && onSegment(p1, q2, q1))
        {
            return true;
        }

        // p2, q2 and p1 are collinear and
        // p1 lies on segment p2q2
        if (o3 == 0 && onSegment(p2, p1, q2))
        {
            return true;
        }

        // p2, q2 and q1 are collinear and
        // q1 lies on segment p2q2
        if (o4 == 0 && onSegment(p2, q1, q2))
        {
            return true;
        }

        // Doesn't fall in any of the above cases
        return false;
}

// Returns true if the point p lies
    // inside the polygon[] with n vertices
function  isInside(polygon,n,p)
{
    // There must be at least 3 vertices in polygon[]
        if (n < 3)
        {
            return false;
        }

        // Create a point for line segment from p to infinite
        let extreme = new Point(INF, p.y);

        // Count intersections of the above line
        // with sides of polygon
        let count = 0, i = 0;
        do
        {
            let next = (i + 1) % n;

            // Check if the line segment from 'p' to
            // 'extreme' intersects with the line
            // segment from 'polygon[i]' to 'polygon[next]'
            if (doIntersect(polygon[i], polygon[next], p, extreme))
            {
                // If the point 'p' is collinear with line
                // segment 'i-next', then check if it lies
                // on segment. If it lies, return true, otherwise false
                if (orientation(polygon[i], p, polygon[next]) == 0)
                {
                    return onSegment(polygon[i], p,
                                    polygon[next]);
                }

                count++;
            }
            i = next;
        } while (i != 0);

        // Return true if count is odd, false otherwise
        return (count % 2 == 1); // Same as (count%2 == 1)
}

// Driver Code
polygon1 = [new Point(0, 0),
                            new Point(10, 0),
                            new Point(10, 10),
                            new Point(0, 10)];

let n = polygon1.length;
        let p = new Point(20, 20);
        if (isInside(polygon1, n, p))
        {
            document.write("Yes<br>");
        }
        else
        {
            document.write("No<br>");
        }
        p = new Point(5, 5);
        if (isInside(polygon1, n, p))
        {
            document.write("Yes<br>");
        }
        else
        {
            document.write("No<br>");
        }
        let polygon2 = [new Point(0, 0),
            new Point(5, 5), new Point(5, 0)];
        p = new Point(3, 3);
        n = polygon2.length;
        if (isInside(polygon2, n, p))
        {
            document.write("Yes<br>");
        }
        else
        {
            document.write("No<br>");
        }
        p = new Point(5, 1);
        if (isInside(polygon2, n, p))
        {
            document.write("Yes<br>");
        }
        else
        {
            document.write("No<br>");
        }
        p = new Point(8, 1);
        if (isInside(polygon2, n, p))
        {
            document.write("Yes<br>");
        }
        else
        {
            document.write("No<br>");
        }
        let polygon3 = [new Point(0, 0),
                            new Point(10, 0),
                            new Point(10, 10),
                            new Point(0, 10)];
        p = new Point(-1, 10);
        n = polygon3.length;
        if (isInside(polygon3, n, p))
        {
            document.write("Yes<br>");
        }
        else
        {
            document.write("No<br>");
        }

// This code is contributed by rag2127
</script>

Output

No 
Yes 
Yes 
Yes 
No 
No 

Time Complexity: O(n) where n is the number of vertices in the given polygon.
Auxiliary Space: O(1), since no extra space has been taken.

This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above