document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1018331>Question 1018331</A>: <I><SPAN STYLE=\"word-break: break-all;\"> One end of diameter of a circle is (-2,-7)and its center is (3,0),find the coordinates of the other end.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #634490 by </B><A HREF=/tutors/aboutme.mpl?userid=mathmate><B>mathmate(429)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mathmate><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=mathmate><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=634490\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  <BR>\n" );
document.write( "Question:<BR>\n" );
document.write( "One end of diameter of a circle is (-2,-7)and its center is (3,0),find the coordinates of the other end.<BR>\n" );
document.write( " <BR>\n" );
document.write( "Solution:<BR>\n" );
document.write( "Since the centre is the mid-point of any diameter, we can use the mid-point formula to solve the problem.<BR>\n" );
document.write( " <BR>\n" );
document.write( "The mid-point formula is:<BR>\n" );
document.write( "(xm,ym)=((x1+x2)/2,(y1+y2)/2)<BR>\n" );
document.write( "where (xm,ym) are the coordinates of the mid point, <BR>\n" );
document.write( "and (x1,y1), (x2,y2) are the coordinates of the two end-points of the diameter.<BR>\n" );
document.write( " <BR>\n" );
document.write( "Therefore<BR>\n" );
document.write( "(x1,y1)=(-2,-7)<BR>\n" );
document.write( "(xm,ym)=(3,0)<BR>\n" );
document.write( "so<BR>\n" );
document.write( "x1=-2<BR>\n" );
document.write( "y1=-7<BR>\n" );
document.write( "xm=3<BR>\n" );
document.write( "ym=0<BR>\n" );
document.write( " <BR>\n" );
document.write( "Applying the formula:<BR>\n" );
document.write( "xm=(x1+x2)/2,<BR>\n" );
document.write( "3=(-2+x2)/2<BR>\n" );
document.write( "(-2+x2)=2*3=6<BR>\n" );
document.write( "x2=6+2=8<BR>\n" );
document.write( " <BR>\n" );
document.write( "ym=(y1+y2)/2<BR>\n" );
document.write( "0=(-7+y2)/2<BR>\n" );
document.write( "-7+y2=0<BR>\n" );
document.write( "y2=7<BR>\n" );
document.write( " <BR>\n" );
document.write( "Therefore the coordinates of the other end is (8,7)<BR>\n" );
document.write( " <BR>\n" );
document.write( " <BR>\n" );
document.write( "Alternative method:<BR>\n" );
document.write( "Since the centre is the mid-point of the diameter, we can use the fact that the change in x and y are the same between one end point and the centre, and the centre to the other end point.<BR>\n" );
document.write( "So<BR>\n" );
document.write( "endpoint=(-2,7)<BR>\n" );
document.write( "centre=(3,0)<BR>\n" );
document.write( "The increase in x from given endpoint to centre = (3-(-2))=5<BR>\n" );
document.write( "The x-coordinate of the other end point is<BR>\n" );
document.write( "3+5=8<BR>\n" );
document.write( " <BR>\n" );
document.write( "The increase in y from given endpoint to centre = (0-(-7))=7<BR>\n" );
document.write( "The y-coordinate of the other end point is<BR>\n" );
document.write( "0+7=7<BR>\n" );
document.write( " <BR>\n" );
document.write( "So the coordinates of the other end point is (8,7)<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );