Ok, I have been working on a Linear Equation program that will draw a line on my graph. But, I am not sure how to set up the y=mx+b in Java to return the (x1,y1) and (x2, y2) using only the slope and y-intercept.

Can anyone tell me how to set up the linear equation in java.

This is a snip bit of my program. First it checks if the user wants to graph something, then it takes the data from the two fields the user entered, [ This part need help on. Getting it to find (x1,y1) and (x2, y2) ], then it converts the answer to points that will match up with the graph lines, draws the data on the graph etc...

If you are simply drawing a line, you can use a MouseListener to pick up the points where the mouse is clicked, and draw a line between the two points. To get m use 1.0 * (y2 - y1) / (x2 - x1). The reason for the 1.0 is to force an implicit cast to double. I can't remember how you get b; I think it would be y1 - x1 * m.

Campbell Ritchie
Sheriff

Joined: Oct 13, 2005
Posts: 44483

34

posted

0

Or you can draw a line between the two points (0, (int)-b) and (w, (int)(-w * m - b)), where w is the width of the component.

The reason for the - is that the y coordinates go from top to bottom.

[edit]Change y to b in two places and add an (int) cast[/edit] [ May 25, 2007: Message edited by: Campbell Ritchie ]

Clarification on problem: I have both m and b all ready. I need to find the other points (x1,y1)(x2,y2) to draw the line.

So, how do I set up the formulas to work in Java to get the other points (x1,y1)(x2,y2) by using m and b? [ May 25, 2007: Message edited by: Emmanuel Xoc ]

Campbell Ritchie
Sheriff

Joined: Oct 13, 2005
Posts: 44483

34

posted

0

I have already told you which formula to try.

Emmanuel Xoc
Greenhorn

Joined: May 24, 2007
Posts: 4

posted

0

I all ready have m and b. I am trying to solve for (x1,y1)(x2,y2).

Could you possibly show an example? I am very lost. y1=b x1=(y-b)/m y2=? x2=? [ May 25, 2007: Message edited by: Emmanuel Xoc ]

You have the slope and the intercept. by definition that means that one point is (0,b).

If all you need is another point, just pick any value for x you want - say, 1. then, stick it into the slope/intercept formula, and calculate y. you now have two points.

example: Slope = 0.2 Intercept = 4

point 1 = ( 0 , 4 )

to find point 2, pick a value for x2, say 1.

i know that y2 = (m)(x2) + b. so, y2 = (0.2) (1) + 4 = 0.2 + 4 = 4.2

point 2 = ( 1 , 4.2 ) [ May 25, 2007: Message edited by: Fred Rosenberger ]

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

Campbell Ritchie
Sheriff

Joined: Oct 13, 2005
Posts: 44483

34

posted

0

Are you sure it is (0, 0.2) Fred?

I would have thought it is (0, 4), only using positive numbers will make the graph display upside down. Agree the next point is (1, 4.2).

You have y = bx + m. You want to put it in a JPanelw pixels wide and h pixels high. You want the "origin" (0, 0) to appear in the centre of the screen. And remember you have to change the sign of the y values because y = 0 is the top of the screen.

Start off with x = 0. The y value is the same as m. Because your y coordinates run downwards, use (w / 2, h / 2 - m). Go to x = 100. You now want to move 100 px to the right, so the first half of the other point is increased by 100, becoming w / 2 + 100. You want to be 100m pixels higher than the origin, so the second part of the point is reduced by 100 * m, becoming h - 2 - m - 100 * m. [You can see the formula can easily be simplified more.]

Originally posted by Campbell Ritchie: [QB]Are you sure it is (0, 0.2) Fred?

Nope. too early in the morning.

thanks. and are YOU sure it's y = bx + m?

I guess the question to the OP is do you need help with the math of y=mx+b, or with the translation of the cartesian coordinates into the JAVA coordinate system?

Campbell Ritchie
Sheriff

Joined: Oct 13, 2005
Posts: 44483

34

posted

0

I think I have got it right this time.

Campbell Ritchie
Sheriff

Joined: Oct 13, 2005
Posts: 44483

34

posted

0

and are YOU sure it's y = bx + m?

It is now!

I think he has got the mathematics right, but needs a lot of help to get it into the proper Cartesian coordinates. [ May 25, 2007: Message edited by: Campbell Ritchie ]

I'm sure its not that difficult to do.. Perhaps using the other form for a straight line would be easier.

y-y1=m(x-x1)

I guess it depends on the JPanel size you are displaying it in, the location of the origin of the graph in the JPanel, and also the scale on the axes, but its really just drawing a line.