document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=65210>Question 65210</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find the equation of a line for this points (4,1) and is perpendicular to the line 2y-3x=4 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #45743 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley71><B>checkley71(8403)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley71><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=45743\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> FIRST WE FIND THE SLOPE OF THE LINE 2Y-3X=4<BR>\n" );
document.write( "2Y=3X-4<BR>\n" );
document.write( "Y=3X/2-4/2<BR>\n" );
document.write( "Y=3X/2-2 THIS LINE HAS A SLOPE OF 3/2 THEREFORE THE PERPENDICULAR LINE HAS A SLOPE OF -2/3.<BR>\n" );
document.write( "THUS THE FORMULAR FOR THIS PERPENDICULAR LINE IS FOUND BY REPLACING X & Y WITH THE POINTS (4,1) WITH A SLOPE OF -2/3 THUS<BR>\n" );
document.write( "1=-2/3*4+b<BR>\n" );
document.write( "1=-8/3+b<BR>\n" );
document.write( "1+8/3=b<BR>\n" );
document.write( "b=11/3 THUS THIS EQUATION IS<BR>\n" );
document.write( "Y=-2/3X+11/3 <BR>\n" );
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document.write( "</TD></TR></TABLE>\n" );