document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=149699>Question 149699</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The vectors (8,0) and (3,4) form a paralellogram. Find the lengths of its altitudes.  I don't know how to generate the algebraic equations to solve this one. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #109810 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=109810\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> First plot the two points (0,0) and (3,4). Now draw a right triangle with the hypotenuse that goes through the two points. With the use of pythagoreans theorem, we find that the hypotenuse is 5 units (since <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%283%5E2%2B4%5E2%29=sqrt%289%2B16%29=sqrt%2825%29=5 ALIGN=MIDDLE  ALT=\"sqrt%283%5E2%2B4%5E2%29=sqrt%289%2B16%29=sqrt%2825%29=5\"   DISABLED_style= \"max-width:90%; height:auto;\" >). So our drawing looks like this:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" src=\"http://i150.photobucket.com/albums/s91/jim_thompson5910/solution2a7-24-08.png\" alt=\"Photobucket - Video and Image Hosting\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Now if draw the vectors and draw the parallelogram, we can see that the height of the parallelogram is the y-coordinate of the point (3,4). So the first height is 4.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" src=\"http://i150.photobucket.com/albums/s91/jim_thompson5910/solution2b7-24-08.png\" alt=\"Photobucket - Video and Image Hosting\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "So from the drawing, we see that the base is 8 units and the height is 4 units.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Now if we compute the area, we get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=b%2Ah=8%2A4=32 ALIGN=MIDDLE  ALT=\"A=b%2Ah=8%2A4=32\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Now let the vector from (0,0) to (3,4) be the new base. So the base is now 5. The area is still 32. So this means that \r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=32=5h ALIGN=MIDDLE  ALT=\"32=5h\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Solving for h, we get <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h=32%2F5=6.4 ALIGN=MIDDLE  ALT=\"h=32%2F5=6.4\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "So the second height is 6.4 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );