document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=270435>Question 270435</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Consider the line is y=3x-6 <BR>\n" );
document.write( "Consider the line  . \r<BR>\n" );
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document.write( "Find the equation of the line that is parallel to this line and passes through the point  .(2,-2)<BR>\n" );
document.write( "Find the equation of the line that is perpendicular to this line and passes through the point  .(2,-2)\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #198128 by </B><A HREF=/tutors/aboutme.mpl?userid=Greenfinch><B>Greenfinch(383)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Greenfinch><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=Greenfinch><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=198128\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The line has a gradient of 3 and a y-intercept of -6<BR>\n" );
document.write( "The || line is also gradient 3. To go through (2,-2) substitute 2 and -2 for x and y so -2 = 2 + c, so c = -4<BR>\n" );
document.write( "Equation is then y = 3x - 4<BR>\n" );
document.write( "The perpendicular gradient is -1/3 since the two gradients multiply to -1.<BR>\n" );
document.write( "So the equation is y = c - 1/3x or 3y + x = c<BR>\n" );
document.write( "To determine c, substitute the values 2 and -2<BR>\n" );
document.write( "-2 = 2/3 + c so c = - 2 2/3<BR>\n" );
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