document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=208331>Question 208331</A>: <I><SPAN STYLE=\"word-break: break-all;\"> write an equation of the line perpendicular to the graph of x-2y-6=0 and passing through A(-3,2) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #157605 by </B><A HREF=/tutors/aboutme.mpl?userid=algebrapro18><B>algebrapro18(249)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=algebrapro18><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=157605\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> write an equation of the line perpendicular to the graph of x-2y-6=0 and passing through A(-3,2)<br>\r<BR>\n" );
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document.write( "Well from this question we have the slope, just not in the current form, and a point that the line will travel through so we will use the point slope formula.  We can also use another method which I will use after I find the answer using the point slope formula.  <br>\r<BR>\n" );
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document.write( "SO we have the equation of a line here in standard form.  To find its slope we need to put it into slope-intercept form(i.e. solve for y).<br>\r<BR>\n" );
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document.write( "x-2y-6=0<BR>\n" );
document.write( "x-2y-6+6=0+6<BR>\n" );
document.write( "x-2y=6<BR>\n" );
document.write( "x-x-2y=6-x<BR>\n" );
document.write( "-2y=6-x<BR>\n" );
document.write( "-2y/-2 = (6-x)/-2<BR>\n" );
document.write( "y = -3+1/2x<BR>\n" );
document.write( "y = 1/2x-3\r<BR>\n" );
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document.write( "so now we know that y = 1/2x-3 and slope intercept form tells us that y = mx+b where m is the slope.  So that means 1/2 is the slope of the origonal line.  Now we need the slope of a line perpenduclar to that line.  That means we use the opposite resiprocal which is -2.<br>\r<BR>\n" );
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document.write( "Therefore we know that our line will have slope -2.  Now we have the slope and the point it passes through, which is (-3,2).  So we can use the point slope formula of y-y1=m(x-x1) where the point (-3,2) are (x1-y1) respectively, m is the slope and x and y are just dummy variables.<br>\r<BR>\n" );
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document.write( "so now we plug in and simplify. <br>\r<BR>\n" );
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document.write( "y-y1=m(x-x1)<BR>\n" );
document.write( "y-2=-2(x--3)<BR>\n" );
document.write( "y-2=-2(x+3)<BR>\n" );
document.write( "y-2+2=-2(x+3)+2<BR>\n" );
document.write( "y = -2(x+3)+2<BR>\n" );
document.write( "y = -2x-6+2<BR>\n" );
document.write( "y = -2x-4<br>\r<BR>\n" );
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document.write( "so the equation of the line perpendicular to the line given going through the point given is y = -2x-4.  <br>\r<BR>\n" );
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document.write( "Now we could also solve this problem another way.  We know that m = -2 and that y = mx+b and we have an x and a y from the point given.  so we can substitute all our knows into the slope-intercept formula and then solve for b.<br>\r<BR>\n" );
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document.write( "y = mx+b<BR>\n" );
document.write( "2 = -2(-3)+b<BR>\n" );
document.write( "2 = 6+b<BR>\n" );
document.write( "-4 = b<br>\r<BR>\n" );
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document.write( "so again we get y = -2x-4 as our answer.  <br> </SPAN>\n" );
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