document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1148023>Question 1148023</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find b such that the points A(2,b),B(5,5)and C(-6,0) are the vertices of right angled <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%28triangle%29 ALIGN=MIDDLE  ALT=\"highlight%28triangle%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > with angle  < BAC = 90 degree.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #769389 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=769389\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> AB distance is sqrt (diff x^2+diff y^2) or sqrt((b-5)^2+9) or sqrt(b^2-10b+34)<BR>\n" );
document.write( "BC distance is sqrt(11^2+5^2)=sqrt(146).  I will call that the hypotenuse<BR>\n" );
document.write( "CA distance is sqrt(8^2+b^2)<BR>\n" );
document.write( "AB^2+CA^2=BC^2<BR>\n" );
document.write( "or b^2-10b+34+b^2+64=146<BR>\n" );
document.write( "or 2b^2-10b+98=146<BR>\n" );
document.write( "or b^2-5b-24=0<BR>\n" );
document.write( "(b-8)(b+3)=0<BR>\n" );
document.write( "b=8 only positive root\r<BR>\n" );
document.write( "\n" );
document.write( "The points are (2, 8) (5, 5) and (-6, 0) </SPAN>\n" );
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