Re: Need help with a problem, is this a place to ask?

Elmer wrote:
> 
> Philippe 92 wrote:
> > ... wrote :
> >> On Tue, 01 Apr 2008 04:19:16 GMT, Elmer <nylicens@[EMAIL PROTECTED]
>
> >> wrote:
> >>> ... wrote:
> >>>> You got a sign wrong.
> >>>>
> >>>>    (-6 - y)^2 = 36 + 12y + y^2
> >>> Well obviously I haven't a clue how to solve 12y+y^2=28.
> >>>
> >>> That's why I kept thinking I was way wrong. Sheest!
> >> Does that mean everything is clear now, or is there still a problem?
> >
> > It seems then that elmer doesn't know how to solve
> > (-6 - y)^2 = 64 ...
> >
> > I have no idea how to *explain* z^2 = 64 => z = +/- 8 ...
> >
> > Regards.
> 
> You're absolutely right. I have no idea what I'm doing. A case of the
blind
> leading the blind.
> 
> d=10 with point A=(4, -6) and B=(-2, y)
> 
> so
> 
> 100=36+(-6-y)^2
> 64=(-6-y)^2
> 
> and here is where everything goes off the rails for me. I haven't a
clue!

What squared equals 64?  Answer 8 or -8.  -6 - y = 8 leads to y = -14,
and -6 - y = -8 leads to y = 2.  If you reflect the segment AB in the
horizontal line through A, you'll see why two answers are possible.

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