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<div class="moz-cite-prefix">On 09/29/14 19:08, wrote:<br>
</div>
<blockquote
cite="mid:C40B2F181831EF44A88CD73525827803130CFB717E@MERCERMAIL.MercerU.local"
type="cite">
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<p>What exactly is alpha in barycentric coordinates, If that is
not needed, how does one figure out the barycentric coordinates
without alpha?</p>
<p><br>
</p>
<p>-Kevin Long<br>
</p>
</blockquote>
<font face="serif"><br>
It is effectively the "third variable" that one has to have to
form the three equations in three unknowns. For example, if you
look on page 26 of the second set of notes, you could rewrite the
matrix as a 3x3<br>
<br>
1 1 1<br>
<br>
0 xb-xc xc-xa<br>
<br>
0 yb-ya yc-ya<br>
<br>
the solution vector be <br>
<br>
alpha<br>
<br>
beta<br>
<br>
gamma<br>
<br>
<br>
and the right vector be<br>
<br>
<br>
1<br>
<br>
xp-xa<br>
<br>
yp-ya<br>
<br>
<br>
<br>
<br>
<br>
</font><br>
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
<a class="moz-txt-link-freetext" href="http://faculty.mercer.edu/pounds_aj">http://faculty.mercer.edu/pounds_aj</a>
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