If-Else Condition in Java: Solving a Quadrant Problem

Introduction

• Today's lecture covers solving the 4th problem on if-else conditions in Java.
• The problem involves determining in which quadrant a point lies based on x and y coordinates provided by the user.

Problem Statement

• Write a program that asks the user to enter two integer values (x and y coordinates).
• Determine the quadrant of the point (x, y) based on the following conditions:
  • Quadrant 1: x > 0 and y > 0
  • Quadrant 2: x < 0 and y > 0
  • Quadrant 3: x < 0 and y < 0
  • Quadrant 4: x > 0 and y < 0
  • Origin: x = 0 and y = 0

Steps to Solve the Problem

Setting Up the Environment

1. Import Scanner:

   import java.util.Scanner;
   

2. Create Scanner Object:

   Scanner sc = new Scanner(System.in);
   

3. Prompt User for Input:
   • For x-coordinate:

     System.out.println("Enter the value of x:");
     int x = sc.nextInt();
     

   • For y-coordinate:

     System.out.println("Enter the value of y:");
     int y = sc.nextInt();
     

Handling Conditions

1. Check for Origin:

   if (x == 0 && y == 0) {
       System.out.println("Origin");
   } else
   

2. Check for Quadrant 1:

   else if (x > 0 && y > 0) {
       System.out.println("Quadrant 1");
   } 
   

3. Check for Quadrant 2:

   else if (x < 0 && y > 0) {
       System.out.println("Quadrant 2");
   } 
   

4. Check for Quadrant 3:

   else if (x < 0 && y < 0) {
       System.out.println("Quadrant 3");
   } 
   

5. Check for Quadrant 4:

   else if (x > 0 && y < 0) {
       System.out.println("Quadrant 4");
   } 
   

Compiling and Running the Program

• Compile the program using a Java compiler.
• Run the program to test different sets of inputs and validate the conditions:
  • Input: x = 8, y = -8 ➔ Output: Quadrant 4
  • Ensure all conditions are handled correctly by modifying input values.

Conclusion

• Handle various conditions to determine the correct quadrant for given x and y values.
• Ensure code structure is clean and all edge cases (like origin) are handled properly.