8 replies to this topic

[atoivan]

1. A PROGRAM THAT PROVIDES THE INTERSECTION COORDINATES OF TWO LINES GIVEN THE COORDINATES OF LINE A AND LINE B.

2. A PROGRAM THAT WILL BE ABLE TO CALCULATE THE GRADIENT/SLOPE OF A LINE GIVEN TWO POINTS.

i want to know what i am doing to do, how to even start it i have no ideas can u pls explain to me how to go about it

NB:
i learn best by doing so i will be glad if u can add example codes to it

[Bat0u89]

Study your theory! In the first question you'll just have to find the functions for line A and line B, say the coordinates for line A are : (x11,y11),(x12,y12)...imagine the first point being somewehre left and down on the cartesian coordinate system and the second righter and upperer :P
(...all you need to define a line is either two points or one point and the line's angle...) then its angle would be a=(y12-y11)/(x12-x11) so its function would be f(x)=ax, the function for line B would look like g(x)=y1-bx (the B line would lean to the left...)...i think it's enough hint

[wim DC]

And for 2:

Where m is the slope.

[atoivan]

wim DC said:

And for 2:

Where m is the slope.

i am trying to put the coordinates of y into an array but it is not working can u correct this for me i do not want to create classes for this program

import javax.swing.JOptionPane;


/**

 *

 * @author irvan

 */

public class Main {

   // private static Object Cord_y;


    /**

     * @param args the command line arguments

     */

    public static void main(String[] args) {

         double  Codr_x[];

         double Codr_y[];

           Codr_y = new double[2];

	    	      Codr_x = new double [2];

                      int z =1;

	    	         for (int i = 0; i < Cord_y.length(); i++) {

	    	          String in =  JOptionPane.showInputDialog("Enter the coordinates for Y"+z);

	    	         Cord_y =Double.parseDouble(in);

                          ++z;

[atoivan]

thanks for the formula
now i am trying to put the coordinates of y into an array but it is not working can u correct this for me i do not want to create classes for this program

import javax.swing.JOptionPane;


/**

 *

 * @author irvan

 */

public class Main {

   // private static Object Cord_y;


    /**

     * @param args the command line arguments

     */

    public static void main(String[] args) {

         double  Codr_x[];

         double Codr_y[];

           Codr_y = new double[2];

	    	      Codr_x = new double [2];

                      int z =1;

	    	         for (int i = 0; i < Cord_y.length(); i++) {

	    	          String in =  JOptionPane.showInputDialog("Enter the coordinates for Y"+z);

	    	         Cord_y =Double.parseDouble(in);

                          ++z;

[wim DC]

Edit: Mr Mike got it in other thread.

[atoivan]

i want the code to be edited for me. and the JOptionPane.showInputDialog is my headache now
i tried mr.mike's correction but the same problem can u do something about it for me.

this is the code with mr.mike's correction u can ran it and see for urself

import javax.swing.JOptionPane;

public class lindemontry1 {


	/**

	 * @param args

	 */

	public static void main(String[] args) {

		// TODO Auto-generated method stub


		  double  Codr_x[];

	         double Codr_y[];

	           Codr_y = new double[2];

		    	      Codr_x = new double [2];

	                      int z =1;

		    	         for (int i = 0; i < Cord_y.length(); i++) {

		    	          String in =  JOptionPane.showInputDialog("Enter the coordinates for Y"+z);

		    	         Cord_y[i] =Double.parseDouble(in);

	                          ++z;

	}


}

Edited by atoivan, 20 May 2011 - 02:36 AM.

[atoivan]

Bat0u89 said:

Study your theory! In the first question you'll just have to find the functions for line A and line B, say the coordinates for line A are : (x11,y11),(x12,y12)...imagine the first point being somewehre left and down on the cartesian coordinate system and the second righter and upperer :P
(...all you need to define a line is either two points or one point and the line's angle...) then its angle would be a=(y12-y11)/(x12-x11) so its function would be f(x)=ax, the function for line B would look like g(x)=y1-bx (the B line would lean to the left...)...i think it's enough hint

thank for the hint i wrote the program i want to know i have done the right thing :

import java.util.Scanner;

public class Linedemo2 {


	/**

	 * @param args

	 */

	public static void main(String[] args) {

		Scanner in =new Scanner(System.in);

		double x1=0,x2=0,x3=0,x4=0,x=0,y=0,y1=0,y2=0,y3=0,y4=0;

		System.out.println("ENTER THE TWO PIONTS FOR LINE A");

		x1 =in.nextDouble();

		y1 =in.nextDouble();

		x2 =in.nextDouble();

		y2 =in.nextDouble();

		System.out.println("ENTER THE TWO PIONTS FOR LINE B");	

		x3 =in.nextDouble();

		y3 =in.nextDouble();

		x4 =in.nextDouble();

		y4 =in.nextDouble();

		double m1,m2,c1,c2;

		m1=(y1-y2)/(x1-x2);//Gradient for line 1

	    m2=(y3-y4)/(x3-x4);//gradient for line 2

	    

	    c1=y1-(m1*x1);  // the y intercept for line 1

	    c2=y3-(m2*x3);   //the y intercept for line 2

	    

	    x=(c1-c2)/(m1-m2);


	    y=(m1*x1)-c1;

	    

	    System.out.println("THE TWO LINE MEET AT POINT(mid-point)" + x  + y );

	    

	}


}

[Bat0u89]

Quote

m1=(y1-y2)/(x1-x2);//Gradient for line 1
m2=(y3-y4)/(x3-x4);//gradient for line 2

c1=y1-(m1*x1); // the y intercept for line 1
c2=y3-(m2*x3); //the y intercept for line 2

x=(c1-c2)/(m1-m2);

y=(m1*x1)-c1;

System.out.println("THE TWO LINE MEET AT POINT(mid-point)" + x + y );

Look, both lines should have a positive gradient (thanks for the proper english word btw :P), suppose the user inputs the line that "leans to the right" first and the one that "leans to the left" second. Suppose too that he inputs the first point of the line first (by first point I mean the left one). Then your code should look like:

m1=(y2-y1)/(x2-x1);//Gradient for line 1
m2=(y3-y4)/(x4-x3);//gradient for line 2
//basically it would look better to get the
//absolute value of each subtraction here...
//Now I show you how to find the x coordinate of the intersection point
//then you can find the y coordinate by m1*x (the function for line 1) OR
// y3-m2*x
//This is simple algebra:
//m1*x=y3-m2*x <=>
//m1*x+m2*x=y3 <=>
//x(m1+m2)=y1 <=>
//x=y1/(m1+m2)
//put all this to code...