points

i have and origin and a destination of a line stored as co ords of a grid (x,y)

i need to calculate if another random point is on the path of that line

any 1 know how?

So I guess you want to find out if your point lies on the line ?

I have a function I made where you give a line, and the an X on the line, and it returns Y coresponding to that X on the line.

You can use the same function for your purposes.

VB Code:

Option Explicit
 
Private Sub Form_Load()
    Dim X1 As Double, X2 As Double, Y1 As Double, Y2 As Double
    Dim TestX As Double, TestY As Double
    
    X1 = 11
    Y1 = 59
    
    X2 = 15
    Y2 = 52
    
    TestX = 31
    TestY = 24
    
    Debug.Print FindYForX(TestX, X1, Y1, X2, Y2) = TestY
End Sub
 
Public Function FindYForX(ByVal X As Double, ByVal X1 As Double, ByVal Y1 As Double, _
        ByVal X2 As Double, ByVal Y2 As Double) As Double
    
    Dim M As Double, B As Double
    
    M = (Y1 - Y2) / (X1 - X2)
    B = Y1 - M * X1
    
    FindYForX = M * X + B
End Function

If points (X1; Y1) and (X2; Y2) belong to a straight line, then any point (x´; y´) will be on the path of this line if you replace x in the equation y = Y1 + [(Y2 – Y1) / (X2 – X1)]*(x – X1) by x´ and you get y´.

Rui

i needed it to work for diagonal lines as well really.

i need to calculate if a given (X,Y) is on a line that spans from (x1,y1) to (x2,y2) i have the following

0= (y2-y1)/(x2-x1)X+ (y1-[(y2-y1)/(x2-x1)]*x1) -Y

but it only seems to work for horizontal lines my maths is pretty bad and i would really appreciate it if any one could explain or even better tell me how i could make it work as i am really lost.

My example should work for any kind of line, horizontal or vertical, it should not be any diference...

Quote:

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Originally Posted by CVMichael

My example should work for any kind of line, horizontal or vertical, it should not be any diference...

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i dont really understand how? so if y = what? when it is on the line

How about this:

VB Code:

Option Explicit
 
Private Sub Form_Load()
 
    Debug.Print Intersects(2, 2, 0, 0, 3, 3)
    Debug.Print Intersects(2, 3, 0, 0, 3, 3)
 
End Sub
 
Private Function Intersects(x As Long, y As Long, x1 As Long, y1 As Long, x2 As Long, y2 As Long) As Boolean
 
    On Error GoTo ERR_Intersects
  
    ' Works out m, and c of y=mx+c, and checks to see if the point appears on the line
    
    Dim m As Long   'Gradient
    Dim c As Long   'Intercept
 
    ' Get m, and c of y=mx+c
    m = (y2 - y1) / (x2 - x1)
    c = y1 - (m * x1)
    
    ' Check against the point for equality
    Intersects = (y = ((m * x) + c))
    
    Exit Function
    
ERR_Intersects:
    Err.Raise Err.Number, Err.Source, Err.Description, Err.HelpFile, Err.HelpContext
End Function

x,y = point of interest
x1,y1 = origin of line
x2,y2 = end of line

Because y=mx+c is the equation of a straight line, and we know enough of the terms we can solve (mx+c) and see if it equals y; if this is true then the y=mx+c holds true, and therefore the point must be on that line.

[edit] you might want to change the datatype into doubles if you want to use this for production code; I haven't adapted it properly - as I'm far too bone idle[/edit]

Quote:

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Originally Posted by nick_p

i dont really understand how? so if y = what? when it is on the line

---

You want to test is X AND Y is on the line right ?

Lets say your X & Y are TestX and TestY
You give to my function TestX, and returns the Y for that TestX
Then you just Test to see if your TestY equals to the returned Y

Just look at my example:

The line coordinates:
X1 = 11
Y1 = 59

X2 = 15
Y2 = 52

The test points X & Y:
TestX = 31
TestY = 24

This line returns Y for TestX, and then tests if the returned Y = TestY....
Debug.Print FindYForX(TestX, X1, Y1, X2, Y2) = TestY

yrwyddfa, that's exactly what my function is doing, except I test the Y outside the function....

Quote:

---

Originally Posted by CVMichael

yrwyddfa, that's exactly what my function is doing, except I test the Y outside the function....

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I understand that I thought the fella was looking for more than a function that solved the problem - my version is annotated to help explain the mathematics behind it.

Thanks guys

sorry abotu that was jsut being stupid i get it now