## math question

Jun 20, 2007 at 11:44pm

# math question

Hi List

I have four points [x,y] on a plane and I need to be able to determine if they are in a line at any angle. I
know that they are in line if all of the x values are equal, or if all of the y values are equal. I do not know
what math to use to check otherwise. Is this a vector problem? trig?

Any hints are appreciated.

Thanks
David

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#32559
Jun 20, 2007 at 11:55pm

One way you could do this is calculate the normal to the plane
that is created by three points. Pick one point, this will be your
origin. Now pick two points and draw a line from them to your origin,
this
should form a plane. Now take your fourth point and create
a vector from the origin, call this the vector x. Now
you can calcualte the angle between the plane and x,
if it is zero then you know all points are in line on the axis
the plane resides on.

Anthony

—– Original Message —–
From: David Morneau
Date: Wednesday, June 20, 2007 6:46 pm
Subject: [maxmsp] math question

> Hi List
>
> I have four points [x,y] on a plane and I need to be able to
> determine if they are in a line at any angle. I
> know that they are in line if all of the x values are equal, or if
> all of the y values are equal. I do not know
> what math to use to check otherwise. Is this a vector problem? trig?
>
> Any hints are appreciated.
>
> Thanks
> David
>
>
>
> Need personalized email and website? Look no further. It’s easy
>

#107421
Jun 21, 2007 at 12:26am

I may have been a bit vague in my explanation. Take a look
at this… http://www.geocities.com/SiliconValley/2151/math3d.html
Once you have calculated the normal to the plane, I think you
can take the dot product of the normal with your remaining
point(vector), that should give you the angle between the two.

Anthony

—– Original Message —–
From: apalomba@austin.rr.com
Date: Wednesday, June 20, 2007 6:57 pm
Subject: Re: [maxmsp] math question
To: maxmsp@cycling74.com, david@5of4.com

> One way you could do this is calculate the normal to the plane
> that is created by three points. Pick one point, this will be your
> origin. Now pick two points and draw a line from them to your
> origin,
> this
> should form a plane. Now take your fourth point and create
> a vector from the origin, call this the vector x. Now
> you can calcualte the angle between the plane and x,
> if it is zero then you know all points are in line on the axis
> the plane resides on.
>
>
>
>
> Anthony
>
>
> —– Original Message —–
> From: David Morneau
> Date: Wednesday, June 20, 2007 6:46 pm
> Subject: [maxmsp] math question
>
> > Hi List
> >
> > I have four points [x,y] on a plane and I need to be able to
> > determine if they are in a line at any angle. I
> > know that they are in line if all of the x values are equal, or
> if
> > all of the y values are equal. I do not know
> > what math to use to check otherwise. Is this a vector problem? trig?
> >
> > Any hints are appreciated.
> >
> > Thanks
> > David
> >
> >
> >
> > Need personalized email and website? Look no further. It’s easy
> >
>

#107422
Jun 21, 2007 at 12:39pm

Take 2 points.
If they are not at the same place, determine the equation of the line they
define (for instance of the form x + a.y + b = 0, the 2 points give you a
system of 2 equations that you solve to find a and b).
Then check if your 3rd and 4th points satisfy the equation.
JF.

> Hi List
>
> I have four points [x,y] on a plane and I need to be able to determine if they
> are in a line at any angle. I
> know that they are in line if all of the x values are equal, or if all of the
> y values are equal. I do not know
> what math to use to check otherwise. Is this a vector problem? trig?
>
> Any hints are appreciated.
>
> Thanks
> David

#107424
Jun 21, 2007 at 12:42pm

Check out the CGA FAQ, subject 1.03, http://www.exaflop.org/docs/cgafaq/.

Good old FAQs.

– Paul

P.S. Here’s the text:
Subject 1.03: How do I find intersections of 2 2D line segments?

This problem can be extremely easy or extremely difficult depends on your
applications. If all you want is the intersection point, the following
should work:

Let A,B,C,D be 2-space position vectors. Then the directed line segments AB
& CD are given by:

AB=A+r(B-A), r in [0,1]
CD=C+s(D-C), s in [0,1]

If AB & CD intersect, then

A+r(B-A)=C+s(D-C), or

Ax+r(Bx-Ax)=Cx+s(Dx-Cx)
Ay+r(By-Ay)=Cy+s(Dy-Cy) for some r,s in [0,1]

Solving the above for r and s yields

(Ay-Cy)(Dx-Cx)-(Ax-Cx)(Dy-Cy)
r = —————————– (eqn 1)
(Bx-Ax)(Dy-Cy)-(By-Ay)(Dx-Cx)

(Ay-Cy)(Bx-Ax)-(Ax-Cx)(By-Ay)
s = —————————– (eqn 2)
(Bx-Ax)(Dy-Cy)-(By-Ay)(Dx-Cx)

Let P be the position vector of the intersection point, then

P=A+r(B-A) or

Px=Ax+r(Bx-Ax)
Py=Ay+r(By-Ay)

By examining the values of r & s, you can also determine some other limiting
conditions:

If 0< =r<=1 & 0<=s<=1, intersection exists
r<0 or r>1 or s<0 or s>1 line segments do not intersect

-

If the denominator in eqn 1 is zero, AB & CD are parallel

-

If the numerator in eqn 1 is also zero, AB & CD are coincident

If the intersection point of the 2 lines are needed (lines in this context
mean infinite lines) regardless whether the two line segments intersect,
then

-

If r>1, P is located on extension of AB
-

If r<0, P is located on extension of BA
-

If s>1, P is located on extension of CD
-

If s<0, P is located on extension of DC

Also note that the denominators of eqn 1 & 2 are identical.

References:

On 6/20/07, Wesley Smith wrote:
> You want to look at slopes i.e.
>
> (y1-y0)/(x1-x0)
>
> if these are the same for all points, then they are colinear.
>
> wes
>
> On 6/20/07,
apalomba@austin.rr.com wrote:
> > I may have been a bit vague in my explanation. Take a look
> > at this… http://www.geocities.com/SiliconValley/2151/math3d.html
> > Once you have calculated the normal to the plane, I think you
> > can take the dot product of the normal with your remaining
> > point(vector), that should give you the angle between the two.
> >
> >
> > Anthony
> >
> >
> > —– Original Message —–
> > From: apalomba@austin.rr.com
> > Date: Wednesday, June 20, 2007 6:57 pm
> > Subject: Re: [maxmsp] math question
> > To: maxmsp@cycling74.com, david@5of4.com
> >
> > > One way you could do this is calculate the normal to the plane
> > > that is created by three points. Pick one point, this will be your
> > > origin. Now pick two points and draw a line from them to your
> > > origin,
> > > this
> > > should form a plane. Now take your fourth point and create
> > > a vector from the origin, call this the vector x. Now
> > > you can calcualte the angle between the plane and x,
> > > if it is zero then you know all points are in line on the axis
> > > the plane resides on.
> > >
> > >
> > >
> > >
> > > Anthony
> > >
> > >
> > > —– Original Message —–
> > > From: David Morneau
> > > Date: Wednesday, June 20, 2007 6:46 pm
> > > Subject: [maxmsp] math question
> > >
> > > > Hi List
> > > >
> > > > I have four points [x,y] on a plane and I need to be able to
> > > > determine if they are in a line at any angle. I
> > > > know that they are in line if all of the x values are equal, or
> > > if
> > > > all of the y values are equal. I do not know
> > > > what math to use to check otherwise. Is this a vector problem? trig?
> > > >
> > > > Any hints are appreciated.
> > > >
> > > > Thanks
> > > > David
> > > >
> > > >
> > > >
> > > > Need personalized email and website? Look no further. It’s easy
> > > > with Doteasy \$0 Web Hosting! Learn more at http://www.doteasy.com
> > > >
> > >
> >
>

—– |(*,+,#,=)(#,=,*,+)(=,#,+,*)(+,*,=,#)| —–

#107425
Jun 21, 2007 at 1:44pm

…or you can use some of the test object from the pmpd library :
http://www.maxobjects.com/?v=libraries&id_library=81
(pmpd.tLine2D, pmpd.tSeg2D, pmpd.tSquare2D… will be your friends)
send it your point position, and the cordinates of the line, segment
or square to test.

Mathieu

http://www.maxobjects.com

http://mathieu.chamagne.free.fr

#107426
Jun 21, 2007 at 1:59pm

Thanks for all of the suggestions. I ended up calculating slopes and comparing, it works great and was
easy to implement.

This whole thing makes me wish I’d paid more attention in my math classes instead of complaining
that I would never need this stuff since I’m in music… oh well.

David

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