document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=980425>Question 980425</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Can someone please solve these 3 coordinate geometry in the x, y plane questions for me? It would really help. E out. Preferably if you show your working out too so I can see how it's done.  \r<BR>\n" );
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document.write( "2) the line PQ is a diameter of the circle center (-4,-2). The line l(small L)  passes through P and is perpendicular to PQ. Given that Q is (10,4), find the equation of l. \r<BR>\n" );
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document.write( "Thank you!!  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #601659 by </B><A HREF=/tutors/aboutme.mpl?userid=Cromlix><B>Cromlix(4381)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Cromlix><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=Cromlix><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=601659\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Hi there,<BR>\n" );
document.write( "First find the coordinates of P<BR>\n" );
document.write( "As Centre of circle is (-4,-2)<BR>\n" );
document.write( "and point Q is (10,4)<BR>\n" );
document.write( "Using 'C' for centre of circle<BR>\n" );
document.write( "Distance QC/Distance CP = 1/1<BR>\n" );
document.write( "c - q /p - c = 1/1<BR>\n" );
document.write( "Cross multiply<BR>\n" );
document.write( "p - c = c - q<BR>\n" );
document.write( "Collect like terms<BR>\n" );
document.write( "p = 2c - q<BR>\n" );
document.write( "p = 2(-4,-2) - (10, 4)<BR>\n" );
document.write( "p = (-18,-8)<BR>\n" );
document.write( "This is the coordinates of P (-18,-8)<BR>\n" );
document.write( "Gradient of CQ (which will equal <BR>\n" );
document.write( "gradient of PQ)<BR>\n" );
document.write( "Gradient = y2 - y1/ x2 - x1<BR>\n" );
document.write( "Using C (-4,-2) and Q(10,4)<BR>\n" );
document.write( "Gradient = 4 - (-2)/10 - (-4)<BR>\n" );
document.write( "Gradient = 6/14 or 3/7<BR>\n" );
document.write( "If line l is perpendicular to PQ<BR>\n" );
document.write( "then their gradients multiply<BR>\n" );
document.write( "together to give -1<BR>\n" );
document.write( "m1 x m2 = -1<BR>\n" );
document.write( "So, line l will have a gradient<BR>\n" );
document.write( "of -7/3<BR>\n" );
document.write( "Using coords of P (-18,-8)<BR>\n" );
document.write( "y - b = m(x - a)<BR>\n" );
document.write( "y - (-8) = -7/3(x - (-18)<BR>\n" );
document.write( "y + 8 = -7/3(x + 18)<BR>\n" );
document.write( "y + 8 = -7/3x - 126/3<BR>\n" );
document.write( "y = -7/3x - 126/3 - 24/3(8)<BR>\n" );
document.write( "y = -7/3x - 50<BR>\n" );
document.write( "Hope this helps:-)<BR>\n" );
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