document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1181250>Question 1181250</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given triangle OAB with vertices O(0, 0), A(5, 2), and B(a, b), let AP and BQ be the perpendiculars dropped from A and B to their opposite sides of triangle OAB ( a not equal to 0, b not equal to 0). Find the constants a and b when the orthocenter of triangle OAB is H(3, 1). \r<BR>\n" );
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document.write( "P.S. Please show full solution, ty.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #811167 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=811167\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "The orthocenter of a triangle is the point of intersection of the three altitudes of the triangle.  The problem will be solved making repeated use of the equations of perpendicular lines.<br><BR>\n" );
document.write( "We have O(0,0) and A(5,2), so the equation of the line containing side OA is y=(2/5)x.<br><BR>\n" );
document.write( "Altitude BQ passes through P(3,1) and is perpendicular to OA, which has slope 2/5; so the line containing altitude BQ has slope -5/2.  Algebra shows BQ is contained in the line y=(-5/2)x+17/2.<br><BR>\n" );
document.write( "Altitude AP passes through A(5,2) and H(3,1); easy algebra shows AP is contained in the line y=(1/2)x-1/2, which has slope 1/2.<br><BR>\n" );
document.write( "Side OB passes through (0,0) and is perpendicular to AP, so its slope is -2; the equation of side OB is then y=-2x.<br><BR>\n" );
document.write( "Point B is the intersection of altitude BQ and side OB:<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-2x+=+%28-5%2F2%29x%2B17%2F2 ALIGN=MIDDLE  ALT=\"-2x+=+%28-5%2F2%29x%2B17%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29x+=+17%2F2 ALIGN=MIDDLE  ALT=\"%281%2F2%29x+=+17%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+17 ALIGN=MIDDLE  ALT=\"x+=+17\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+-2x+=+-2%2817%29=-34 ALIGN=MIDDLE  ALT=\"y+=+-2x+=+-2%2817%29=-34\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "ANSWER: B(17,-34)<br><BR>\n" );
document.write( "To confirm our method and our calculations, we can show that the altitude from O to side AB gives an equation for the line containing side AB that gives us the same coordinates for B.<br><BR>\n" );
document.write( "The slope of the altitude from O is 1/3, so the slope of AB is -3.<BR>\n" );
document.write( "AB with a slope of -3 and passing through A(5,2) gives the equation y=-3x+17 for the line containing side AB.<BR>\n" );
document.write( "Substitution verifies that (17,-34) is on that line.<br><BR>\n" );
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