> J is a language concerned with problems MUCH more complex than yours.

This is a misconception.



> From: James C Field <[email protected]>
> 
> I suggest you explore the internet and refer to printed versions of 
> basic geometry. The answer to your problem can be solved with PAPER and 
> PENCIL.
> 
> J is a language concerned with problems MUCH more complex than yours.
> 
> First... Draw a pencil diagram of the problem.
> Second ... Draw a sketch reconsidering the initial parameters of the 
> problem.
> Third...stop and think and the go to bed .... think again in the 
> morning. IT IS AMAZING HOW SMART YOU CAN BE IN THE MORNING!
> 
> Richard Donovan wrote:
> > Point equidistant to 3 or more other points  
> >             
> >             Hi!
> >  
> >
> > Not specifically related to J, a maths question really, but I will try to 
> > code 
> the solution in J!
> >  
> >
> > If there are 2 points, (x1 y1) and (x2 y2), the equidistant point is at the 
> midpoint at location ((x1+x2)/2) ((y1+y2)/2)
> >  
> >
> > eg if there are 2 points p and q at locations (1,2) and (3,0) the point 
> equidistant is at (2,1) - the midpoint m
> >  
> >
> >    y 
> >
> > 4 |
> >
> > 3 |
> >
> > 2 |p
> >
> > 1 |    m
> >
> > _ |__q_______ x
> >
> >   0 1 2 3 4
> >  
> >
> > Suppose there are more than 2 points? Is there a formula for working out 
> > the 
> point equidistant to the n points?
> >  
> >
> > eg if there are 3 points p q r at locations (1,2) (3,0) and (3,2), the
> > point equidistant is still at (2,1), but I can't see a way of computing
> > this from the figures.
> >  
> >
> >    y 
> >
> > 4 |
> >
> > 3 |
> >
> > 2 |p--r
> >
> > 1 |    m
> >
> > _ |__q_______ x
> >
> >   0 1 2 3 4                  NB hyphens are spaces-hotmail changes 2 spaces 
> > to 
> 1
> >  
> >
> > This is not a test question or homework crib - I am just interested in 
> mathematics!
> >  
> >
> > The thought occured to me and I just cant see a solution for the
> > general case finding equidistant point of points (x1 y1) (x2 y2) (x3 y3) ...
> > (xn yn)
> >
> > ...and as for the 3D case (x1 y1 z1) (x2 y2 z2) (x3 y3 z3) ...
> > (xn yn zn)......!!!!!!
> >  
> >
> > Thanks in advance!
> >
> >             
> >
> >             
> >         
> >                                
> > _________________________________________________________________
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> >
> >  
> 
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