CSES Solutions – Point Location Test

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:

• If the cross product is positive, p₃ is on the left side of the line.
• If the cross product is negative, p₃ is on the right side of the line.
• If the cross product is zero, p₃ is on the line.

Steps to solve the problem:

• Calculate the cross product using the formula: (x2 – x1) * (y3 – y1) – (y2 – y1) * (x3 – x1).
• If the cross product is less than 0, then point p₃ is on the right side of the line passing through p₁ and p₂. Print “RIGHT“.
• If the cross product is greater than 0, then point p₃ is on the left side of the line passing through p₁ and p₂. Print “LEFT“.
• If the cross product is equal to 0, then point p₃ lies on the line passing through p₁ and p₂. Print “TOUCH“.

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();

Output

LEFT

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