document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=347468>Question 347468</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The triangle inequality states: PQ + QR > PR<BR>\n" );
document.write( "Prove that the equality holds if and only if the point Q lies on the line segment PR when we use the usual Euclidean distance formula. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #248658 by </B><A HREF=/tutors/aboutme.mpl?userid=nyc_function><B>nyc_function(2741)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=nyc_function><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=nyc_function><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=248658\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Without loss of generality, we can place P at (0,0), R at (a,0), and Q at (x,y)<BR>\n" );
document.write( "with a>0<BR>\n" );
document.write( "then the distances are<BR>\n" );
document.write( "PQ: sqrt( x^2 + y^2 )<BR>\n" );
document.write( "QR: sqrt( (x-a)^2 + y^2 )<BR>\n" );
document.write( "PR: a<BR>\n" );
document.write( "thus we want to know when<BR>\n" );
document.write( "PQ + QR = PR or<BR>\n" );
document.write( "sqrt( x^2 + y^2 ) + sqrt( (x-a)^2 + y^2 ) = a squaring both sides we get<BR>\n" );
document.write( "x^2 + y^2 + 2*sqrt[ (x^2+y^2) * ( (x-a)^2 + y^2) ] + (x-a)^2 + y^2 = a^2<BR>\n" );
document.write( "2*sqrt[ (x^2+y^2) * ( (x-a)^2 + y^2 ) ] = 2ax - 2x^2 - 2y^2<BR>\n" );
document.write( "(x^2 + y^2) * ( (x-a)^2 + y^2 ) = (ax - x^2 - y^2) ^2<BR>\n" );
document.write( "(x^2+y^2)^2 - 2ax(x^2+y^2) + a^2(x^2+y^2) = a^2x^2 - 2ax(x^2+y^2) + (x^2+y^2)<BR>\n" );
document.write( "a^2(x^2+y^2) = a^2x^2 since a>0 we can divide it out<BR>\n" );
document.write( "x^2+y^2 = x^2<BR>\n" );
document.write( "y^2 = 0<BR>\n" );
document.write( "y=0<BR>\n" );
document.write( "now plugging this into original equation we get<BR>\n" );
document.write( "|x| + |x-a| = a<BR>\n" );
document.write( "now if x<0 then we get<BR>\n" );
document.write( "-x + a -x = a<BR>\n" );
document.write( "-2x = 0<BR>\n" );
document.write( "x=0 which contradicts x<0 thus x>=0<BR>\n" );
document.write( "if x>a then we get<BR>\n" );
document.write( "x + x -a = a<BR>\n" );
document.write( "2x = 2a<BR>\n" );
document.write( "x=a which contradicts x>a<BR>\n" );
document.write( "thus 0<=x<=a<BR>\n" );
document.write( "thus point for PQ+QR=PR to be true, Q must be on the line segment PR<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );