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CSES Solutions - Point Location Test

CSES Solutions - Point Location Test

Last Updated : 23 Jul, 2025

There is a line that goes through the points p1 = (x1,y1) and p2 = (x2,y2). There is also a point p3 = (x3,y3).

Your task is to determine whether p3 is located on the left or right side of the line or if it touches the line when we are looking from p1 to p2.

Example:

Input: x1 = 1, y1 = 1, x2 = 5, y2 = 3, x3 = 2, y3 = 3
Output: LEFT

Input: x1 = 1, y1 = 1, x2 = 5, y2 = 3, x3 = 4, y3 = 1
Output: RIGHT

Approach:

The cross product of two vectors a and b is a vector that is perpendicular to both a and b and thus normal to the plane containing them. The direction of the cross product vector is given by the right-hand rule, and its magnitude is equal to the area of the parallelogram that the vectors span.

But for this problem, we are not interested in the resulting vector itself, but rather the sign of its z-component (since we are working in 2D space, this is the only component of the cross product). The sign of the z-component of the cross product tells us whether p3 is to the left or right of the line from p1 to p2:

Steps to solve the problem:

Below are the implementation of the above approach:

C++ 
#include <bits/stdc++.h>
using namespace std;

int main()
{

    // Coordinates of points p1, p2, and p3
    long long x1 = 1, y1 = 1, x2 = 5, y2 = 3, x3 = 2, y3 = 3;

    // Calculate the cross product to determine the position of p3 relative to the line through p1 and p2
    long long crossProduct = (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1);

    // If cross product is less than 0, p3 is on the right of the line
    if (crossProduct < 0)
        cout << "RIGHT" << endl;
    // If cross product is more than 0, p3 is on the left of the line
    else if (crossProduct > 0)
        cout << "LEFT" << endl;
    // If cross product is 0, p3 is on the line
    else
        cout << "TOUCH" << endl;

    return 0;
}
Java 
import java.util.*;

public class Main {
    public static void main(String[] args) {
        // Coordinates of points p1, p2, and p3
        long x1 = 1, y1 = 1, x2 = 5, y2 = 3, x3 = 2, y3 = 3;

        // Calculate the cross product to determine the position of p3 relative to the line through p1 and p2
        long crossProduct = (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1);

        // If cross product is less than 0, p3 is on the right of the line
        if (crossProduct < 0)
            System.out.println("RIGHT");
        // If cross product is more than 0, p3 is on the left of the line
        else if (crossProduct > 0)
            System.out.println("LEFT");
        // If cross product is 0, p3 is on the line
        else
            System.out.println("TOUCH");
    }
}
Python3 
# Python code to determine the position of a point relative to a line

# Coordinates of points p1, p2, and p3
x1, y1 = 1, 1
x2, y2 = 5, 3
x3, y3 = 2, 3

# Calculate the cross product to determine the position of p3 relative to the line through p1 and p2
crossProduct = (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1)

# If cross product is less than 0, p3 is on the right of the line
if crossProduct < 0:
    print("RIGHT")
# If cross product is more than 0, p3 is on the left of the line
elif crossProduct > 0:
    print("LEFT")
# If cross product is 0, p3 is on the line
else:
    print("TOUCH")
JavaScript 
function main() {
    // Coordinates of points p1, p2, and p3
    let x1 = 1, y1 = 1, x2 = 5, y2 = 3, x3 = 2, y3 = 3;

    // Calculate the cross product to determine the position of p3 relative to the line through p1 and p2
    let crossProduct = (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1);

    // If cross product is less than 0, p3 is on the right of the line
    if (crossProduct < 0)
        console.log("RIGHT");
    // If cross product is more than 0, p3 is on the left of the line
    else if (crossProduct > 0)
        console.log("LEFT");
    // If cross product is 0, p3 is on the line
    else
        console.log("TOUCH");
}

// Call the main function
main();

Time Complexity: O(1)
Auxiliary Space: O(1)


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