Math Problem!

[mrnoob]

On a Cartesian plane, there is a point (x,y). There is a second point (i,n). Find the point on the line which passes through (x,y) and (i,n) which has the largest value of x which is less then or equal to z, or which has the largest value of y which is less then or equal to 2z.

Pictorial example - http://img505.imageshack.us/my.php?image=gaymd0.jpg

Edited December 5, 2007 by mrnoob

[weaponx]

What are you talking about? You can't just say "a point". A point in space? A point on a graph? A point on an image? Tell us what you are trying to accomplish.

[mrnoob]

A point on a graph.

I am trying move a cursor starting at one point, then passing through a second point and continuing on until either x>=932 or y >=370

Edited December 5, 2007 by mrnoob

[nikink]

That would just require some kind of conditional, wouldn't it?

If $x >= 932 Then

Return $y

ElseIf $y >=370 Then

Return $x

EndIf

?

[mrnoob]

Yes, thats helpful, but I need a way to find a point which is on the line that passes through the first and second point

This may help, http://img255.imageshack.us/my.php?image=thingot2.jpg.

[weaponx]

There isn't so much a formula as there is a condition, since there is many many x,y values that match this criteria.

If ($x >= 932 or $y >=370) Then

;Stop

EndIf

- or -

Do

;code

Until ($x >= 932 or $y >=370)

[mrnoob]

There is only one point that will satisfy what I am looking for; there are not many x,y coords that will match this criteria.

[Nahuel]

Given two points, you can easily find the equation of a rect. What you are trying to do, If I'm not wrong, is make the mouse move in a line which contains these previously given coordinates and continue in a straight line unitl x >= 932 or y >=370

If that's so, find the equation that contains your two origian points. In a loop, increase X in the equation, this will give you Y. You now have a new (x,y) point to make the mouse move to. Do this until x >= 932 or y >=370.

[Xandl]

Hello,

this is the formula for a line through two points:

(y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)

So, we can get

x = ((y - y1)/(y2 - y1) * (x2 - x1)) + x1

and

y = ((x - x1)/(x2 - x1) * (y2 - y1)) + y1

x is what you are looking for, y will be yMin and yMax in your plane.

The same goes with y.

This example shows a correct result:

ConsoleWrite(lX(0,0,10,10,20)&@LF)

Func lX($x1,$y1,$x2,$y2,$maxY)
    Return (($maxY - $y1)/($y2 - $y1) * ($x2 - $x1)) + $x1
EndFunc

ciao

Xandl

[Nahuel]

Here's an example of what I meant:

Global $X
Global $Y
;Point 1:
Global $X_1 = 5
Global $Y_1 = 10

;Point 2:
Global $X_2 = 100
Global $Y_2 = 100

$X=$X_1
$Y=$Y_1
$Past=False;This is just to show the messagebox once.
Do
    MouseMove($X,$Y,2000)
    If $X>=$X_2 And $Y>=$Y_2 And not $Past Then 
        MsgBox(0,"","We just past by point 2!")
        $Past=True
    EndIf
;Increase X
    $X+=5; The higher the faster.
    
;Equation of a rect given two points:
    $Y = (($Y_2 - $Y_1)/($X_2 - $X_1))*($X-$X_1) + $Y_1
    
    $Y=Int($Y)
    ConsoleWrite("X: " & $X & ", Y: " & $Y & @CR)
Until ($X >= 932 or $Y >=370)

-edit-

@Xandl: Precisely

Edited December 5, 2007 by Nahuel

[mrnoob]

Ah, thank you guys so much; (y - y1)/(y2 - y1) = (x - x1)/(x2 - x1) is exactly what I was looking for!!!! I'm curious if you figured that out or if its a common formula?

[weaponx]

Finding the slope. http://en.wikipedia.org/wiki/Slope