document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=893194>Question 893194</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the equation of the circle inscribed in a triangle, if the triangle has its sides on the lines 2x+y-9=0, -2x+y-1=0, -x+2y+7=0. Draw the figure  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #541155 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=541155\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find the equation of the circle inscribed in a triangle, if the triangle has its sides on the lines 2x+y-9=0, -2x+y-1=0, -x+2y+7=0. Draw the figure <BR>\n" );
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document.write( "Idk any shortcuts, just have to grind it out.<BR>\n" );
document.write( "Find the intersections, then the lengths of the sides:<BR>\n" );
document.write( "---<BR>\n" );
document.write( "2x +y-9=0<BR>\n" );
document.write( "-2x+y-1=0<BR>\n" );
document.write( "------------- Add<BR>\n" );
document.write( "2y -10 = 0<BR>\n" );
document.write( "y = 5<BR>\n" );
document.write( "x = 2<BR>\n" );
document.write( "Point A(2,5)<BR>\n" );
document.write( "----<BR>\n" );
document.write( "-2x+y-1=0<BR>\n" );
document.write( "-x+2y+7=0<BR>\n" );
document.write( "Point B(-3,-5)<BR>\n" );
document.write( "----<BR>\n" );
document.write( "2x+y-9=0<BR>\n" );
document.write( "-x+2y+7=0<BR>\n" );
document.write( "Point C(5,-1)<BR>\n" );
document.write( " <BR>\n" );
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document.write( "Find the lengths AB, BC and AC<BR>\n" );
document.write( "side c = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AB+=+sqrt%28diffy%5E2+%2B+diffx%5E2%29+=+sqrt%285%5E2+%2B+10%5E2%29 ALIGN=MIDDLE  ALT=\"AB+=+sqrt%28diffy%5E2+%2B+diffx%5E2%29+=+sqrt%285%5E2+%2B+10%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "c = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AB+=+5sqrt%285%29 ALIGN=MIDDLE  ALT=\"AB+=+5sqrt%285%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "b = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AC+=+3sqrt%285%29 ALIGN=MIDDLE  ALT=\"AC+=+3sqrt%285%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "a = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=BC+=+4sqrt%285%29 ALIGN=MIDDLE  ALT=\"BC+=+4sqrt%285%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "It's a right triangle, btw.  Sides of 3, 4 & 5 times sqrt(5)<BR>\n" );
document.write( "----<BR>\n" );
document.write( "The center of the circle is on the intersection of the bisectors of the 3 angles.<BR>\n" );
document.write( "The intersection of any 2 bisectors is the center.<BR>\n" );
document.write( "Find the slope of the 3 sides:<BR>\n" );
document.write( "2x+y-9=0, -2x+y-1=0, -x+2y+7=0<BR>\n" );
document.write( "Put each of the 3 original eqns in slope-intercept form.<BR>\n" );
document.write( "2x+y-9=0<BR>\n" );
document.write( "y = -2x + 9<BR>\n" );
document.write( "m = -2 (slope of AC)<BR>\n" );
document.write( "----<BR>\n" );
document.write( "-2x+y-1=0<BR>\n" );
document.write( "y = 2x + 1<BR>\n" );
document.write( "m = 2 (slope of BA)<BR>\n" );
document.write( "----<BR>\n" );
document.write( "-x+2y+7=0<BR>\n" );
document.write( "y = x/2 - 7/2<BR>\n" );
document.write( "m = 1/2 (slope of BC)<BR>\n" );
document.write( "========================<BR>\n" );
document.write( "The slopes are AC and BA are opposites, so the bisector of A is vertical (parallel to the y-axis).<BR>\n" );
document.write( "--> the center of the circle is on the line x = 2.<BR>\n" );
document.write( "==============<BR>\n" );
document.write( "Using r = a*b/(a+b+c), the radius of the circle = sqrt(5)<BR>\n" );
document.write( "---<BR>\n" );
document.write( "The intersection of the bisector of A and side BC is (2,-2.5) (point D)<BR>\n" );
document.write( "Distance DC = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3sqrt%285%29%2F2%29 ALIGN=MIDDLE  ALT=\"3sqrt%285%29%2F2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "---------<BR>\n" );
document.write( "Triangles ACD and AEO are similar, so<BR>\n" );
document.write( "AD/DC = AO/OE<BR>\n" );
document.write( "AD = 15/2<BR>\n" );
document.write( "7.5/(3sqrt(5)/2) = AO/sqrt(5)<BR>\n" );
document.write( "AO = 5<BR>\n" );
document.write( "===========<BR>\n" );
document.write( "--> The center is (2,0)<BR>\n" );
document.write( "---------<BR>\n" );
document.write( "The circle is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x-2%29%5E2+%2B+y%5E2+=+5 ALIGN=MIDDLE  ALT=\"%28x-2%29%5E2+%2B+y%5E2+=+5\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "======================<BR>\n" );
document.write( "I'll send a graph if you respond via the TY note.\r<BR>\n" );
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