document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=35483>Question 35483</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi, not sure if this is appropriate topic, anyway here goes<BR>\n" );
document.write( "for the point A (3,-1) and the line with equation x-y=1, find:\r<BR>\n" );
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document.write( "a   The equation of the perpendicular to the line through A<BR>\n" );
document.write( "b   The foot of the perpendicular from A to the line<BR>\n" );
document.write( "c   The distance of A from the line\r<BR>\n" );
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document.write( "Hope someone can help, thank you </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #21629 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=21629\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Hi, not sure if this is appropriate topic, anyway here goes<BR>\n" );
document.write( "for the point A (3,-1) and the line with equation x-y=1....SAY LINE L<BR>\n" );
document.write( " find:<BR>\n" );
document.write( "a The equation of the perpendicular to the line through A<BR>\n" );
document.write( "EQN.OF LINE PERPENDICULAR TO L IS OF THE FORM <BR>\n" );
document.write( "X+Y=K(..RULE IS EXCHANGE COEFFICIENTS OF X AND Y AND MAKE ONE NEGATIVE AND EQUATE THE RESULT TO A CONSTANT TO BE FOUND OUT) THIS IS TO GO THROUGH A ..HENCE..<BR>\n" );
document.write( "3+(-1)=K=2..........HENCE EQN. OF REQD.LINE IS <BR>\n" );
document.write( "X+Y=2<BR>\n" );
document.write( "b The foot of the perpendicular from A to the line<BR>\n" );
document.write( "LET B(P,Q) BE THE FOOT OF PERPENDICULAR FROM A ON TO THE LINE.<BR>\n" );
document.write( "SLOPE OF AB =(Q+1)/(P-3)...SINCE THIS IS PERPENDICULAR TO OUR LINE SLOPE =1.HENCE<BR>\n" );
document.write( "(Q+1)/(P-3)=1...............OR.....Q+1=P-3<BR>\n" );
document.write( "P-Q=4.....................I<BR>\n" );
document.write( "BUT B IS ON LINE ...SO....<BR>\n" );
document.write( "P+Q=2.........................II<BR>\n" );
document.write( "ADDING EQN.I AND EQN.II<BR>\n" );
document.write( "2P=6<BR>\n" );
document.write( "P=3<BR>\n" );
document.write( "Q=2-P=2-3=-1<BR>\n" );
document.write( "HENCE FOOT OF PERPENDICULAR IS (3,-1)\r<BR>\n" );
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document.write( "c The distance of A from the line<BR>\n" );
document.write( "IT IS EQUAL TO AB .HENCE <BR>\n" );
document.write( "AB=SQRT{(3-3)^2+(1+1)^2}=SQRT(2^2)=2\r<BR>\n" );
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document.write( "Hope someone can help, thank you </SPAN>\n" );
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