document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=34632>Question 34632</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0321227573/algebrahomeworkh>College algebra</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> Question; Write and equation (a) in standard form and (b) in slope-intercept form.  # 38. Through(-2,0), perpendicular to 8x-3y=7<BR>\n" );
document.write( "I really don't understand what \"Through\" means? I don't understand the equation either. But I really don't get through. Thanks so very much! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #21187 by </B><A HREF=/tutors/aboutme.mpl?userid=kietra><B>kietra(57)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=kietra><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=21187\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> There is a lot to this question, so lets first start with parts a and b.  I am not sure if you have to have a specific or general equation.  Standard form of an equation is ax + by = c.  You can have any numbers for a,b,and c.  Slope intercept form is y=mx+b and m and b could be any numbers.  For #38- it wants you to find an equation that goes through the point (-2,0) and is perpendicular to 8x-3y=7.  I hope that makes sense- maybe it will make more as we keep going.  Lets start with 8x-3y=7.  In order for two lines to be perpendicular, the product of their slopes has to be -1, or they have to be inverses.  Lets first find the slope of 8x-3y=7.  Put that equation in slope intercept form: <BR>\n" );
document.write( "8x - 3y = 7<BR>\n" );
document.write( "-3y = -8x + 7<BR>\n" );
document.write( "y = 8/3x - 7/3<BR>\n" );
document.write( "The only important number in this equation is 8/3 - that is the slope of this equation.  So, the slope of a line perpendicular to that would be its inverse or -3/8.  \r<BR>\n" );
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document.write( "Now, we have the slope of the line and a point that this line goes through (-2,0).   Lets use point slope form (because we know a point and the slope) to find the equation of the line.  POint slope is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m=%28y-y1%29%2F%28x-x1%29 ALIGN=MIDDLE  ALT=\"m=%28y-y1%29%2F%28x-x1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >where (x1,y1) is the point on our line. <BR>\n" );
document.write( "Lets substitute: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-3%2F8%29=%28y+-+0%29%2F%28x+-+%28-2%29%29 ALIGN=MIDDLE  ALT=\"%28-3%2F8%29=%28y+-+0%29%2F%28x+-+%28-2%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Now solve for y to get it into slope intercept:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-3%2F8%29=%28y+-+0%29%2F%28x+-+%28-2%29%29 ALIGN=MIDDLE  ALT=\"%28-3%2F8%29=%28y+-+0%29%2F%28x+-+%28-2%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-3%2F8%29=%28y%29%2F%28x+%2B+2%29%29 ALIGN=MIDDLE  ALT=\"%28-3%2F8%29=%28y%29%2F%28x+%2B+2%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-3%2F8%29%2A%28x+%2B+2%29=y ALIGN=MIDDLE  ALT=\"%28-3%2F8%29%2A%28x+%2B+2%29=y\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-3%2F8%29%2Ax+%2B+%28-3%2F8%29%2A2+=+y ALIGN=MIDDLE  ALT=\"%28-3%2F8%29%2Ax+%2B+%28-3%2F8%29%2A2+=+y\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-3%2F8%29%2Ax+%2B+%28-3%2F4%29+=+y ALIGN=MIDDLE  ALT=\"%28-3%2F8%29%2Ax+%2B+%28-3%2F4%29+=+y\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Now, to get in standard form, put x back on the same side with y and get rid of your fraction:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-3%2F8%29%2Ax+%2B+%28-3%2F4%29+=+y ALIGN=MIDDLE  ALT=\"%28-3%2F8%29%2Ax+%2B+%28-3%2F4%29+=+y\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-3%2F4%29+=+y+%2B+%283%2F8%29%2Ax ALIGN=MIDDLE  ALT=\"%28-3%2F4%29+=+y+%2B+%283%2F8%29%2Ax\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Multiply everything by 8 to get rid of fractions<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=8%2A%28-3%2F4%29+=8y+%2B+8%283%2F8%29%2Ax ALIGN=MIDDLE  ALT=\"8%2A%28-3%2F4%29+=8y+%2B+8%283%2F8%29%2Ax\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-6=+8y+%2B+3x ALIGN=MIDDLE  ALT=\"-6=+8y+%2B+3x\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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