document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=96947>Question 96947</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the straight line y= 2x + 3 meets the curve xy+20=5y at the points A and B.Find the equation of the perpendicular bisector of AB. i cant just cant seem to work this out </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #70598 by </B><A HREF=/tutors/aboutme.mpl?userid=mathslover><B>mathslover(157)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mathslover><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=mathslover><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=70598\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> First we need to find the intersection points of the curve xy + 20 =5y and the straight line y= 2x + 3 to get the points A and B\r<BR>\n" );
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document.write( "Substituting y= 2x+3 in the equation of the curve we have,\r<BR>\n" );
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document.write( "x(2x + 3) + 20 = 5(2x + 3) \r<BR>\n" );
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document.write( "2x^2 + 3x + 20 = 10x +15\r<BR>\n" );
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document.write( "2x^2 -7x + 5 =0\r<BR>\n" );
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document.write( "Using the quadratic formula to get the values of x\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ ALIGN=MIDDLE  ALT=\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" > \r<BR>\n" );
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document.write( "x =  7 +- (sqrt(49 -40))/(4) <BR>\n" );
document.write( "x=5/2 , 1\r<BR>\n" );
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document.write( "putting these values of x in the equation y = 2x + 3<BR>\n" );
document.write( "y = 8 and y = 5\r<BR>\n" );
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document.write( "Therefore the points of intersection of the straight line and the curve is <BR>\n" );
document.write( "(5/2,8) and (1,5)\r<BR>\n" );
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document.write( "The perpendicular bisector of AB divides the line segment AB in equal halves and therefore if D is the point of bisection the co-ordinates of D are \r<BR>\n" );
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document.write( "((5/2 +1)/2 , (8 + 5)/2  )  Using the mid point formula <BR>\n" );
document.write( "((x1 + x2)/2, (y1+ y2)/2 )<BR>\n" );
document.write( "= (7/4, 13/2)\r<BR>\n" );
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document.write( "we are left with finding the equation of the line through (7/4, 13/2) and perpendicular to the line  y =2x +3 \r<BR>\n" );
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document.write( "slope of the line y=2x +3  is 2  ( y=mx +c where m is the slope )<BR>\n" );
document.write( "if p is the slope of the perpendicular than p* 2 = -1 (since product of the slope of a line and its perpendicular is -1)\r<BR>\n" );
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document.write( "therefore p =-1/2<BR>\n" );
document.write( "so if we represent y=mx + c as the equation of the perpendicular \r<BR>\n" );
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document.write( "we have <BR>\n" );
document.write( "13/2= -1/2 * 7/4 + c ( Substituting values of (x,y) and m )<BR>\n" );
document.write( "C= 13/2 + 7/8<BR>\n" );
document.write( "c = 59/8\r<BR>\n" );
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document.write( "Hence equation of the line is y = -1/2x + 59/8<BR>\n" );
document.write( "Multiplying by 8 on both sides\r<BR>\n" );
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document.write( "8y = -4x + 59<BR>\n" );
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