(6 , 4) is outside the rectangle, as shown in figure below. (Hint: A point is in the rectangle if its horizontal distance to (0, 0) is less than or equal to 10/2 and its vertical distance to (0, 0) is less than or equal to 5.0/2.)
Solution:
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
double x, y;
System.out.print("Enter the value of x and y:");
x = s.nextDouble();
y = s.nextDouble();
if(x < 0) {
x = x * (-1);
}
if(y < 0) {
y = y * (-1);
}
if((x < 10.0/2) && (y < 5.0/2)) {
System.out.print("Point (" + x + ", " + y + ") is in the rectangle.");
} else {
System.out.print("Point (" + x + ", " + y + ") is not in the rectangle.");
}
}
Scanner input = new Scanner(System.in);
ReplyDeleteSystem.out.print("Enter a point with two coordinates:");
double x = input.nextDouble();
double y = input.nextDouble();
String s = " ";
if (Math.abs(x) > 5 || Math.abs(y) > 2.5) {
s = " not ";
}
System.out.print("Point (" + x + ", " + y + ") is" + s
+ "in the rectangle");
import java.util.Scanner;
ReplyDeletepublic class Exercise_03_23 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
// Prompt the user to enter a point (x, y)
System.out.print("Enter a point with two coordinates: ");
double x = input.nextDouble();
double y = input.nextDouble();
// Check whether the point is within the rectangle
// centered at (0, 0) with width 10 and height 5
boolean withinRectangle = (Math.pow(Math.pow(x, 2), 0.5) <= 10 / 2 ) ||
(Math.pow(Math.pow(y, 2), 0.5) <= 5.0 / 2);
// Display results
System.out.println("Point (" + x + ", " + y + ") is " +
((withinRectangle) ? "in " : "not in ") + "the rectangle");
}
}
can you make this as a flowchart?
ReplyDelete