document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=864936>Question 864936</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find the point whose distance from (7,-3) is square root of 58 and whose abscissa equals its ordinate. Please show your solution. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #521397 by </B><A HREF=/tutors/aboutme.mpl?userid=richwmiller><B>richwmiller(17219)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richwmiller><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=richwmiller><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=521397\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the point will be (y,y) or (x,x) since the abscissa = the ordinate x=y<BR>\n" );
document.write( "sqrt(58)<BR>\n" );
document.write( "58=(7-x)^2+(-3-x)^2\r<BR>\n" );
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document.write( "(0,0) and (4,4) are sqrt(58) from (7,-3) </SPAN>\n" );
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