document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=199381>Question 199381</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Analytic Geometry<BR>\n" );
document.write( "14. Determine the shortest distance from the point H(5,2) tot he line through points J(-6,4) and K(-2,-4)\r<BR>\n" );
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document.write( "Thank you alot!! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #149816 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=149816\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Determine the shortest distance from the point H(5,2) tot he line through points J(-6,4) and K(-2,-4) <BR>\n" );
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document.write( "Method:  Find the slope of the line JK.  The shortest distance from a point to a line is along a line perpendicular, to JK in this case.  The slope of lines perpendicular is the negative inverse of the slope of JK.<BR>\n" );
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document.write( "Slope, m, of JK = diffy/diffx = (-4-4)/(-2 - (-6)) = -8/4<BR>\n" );
document.write( "m of JK = -2<BR>\n" );
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document.write( "The slope of ALL lines perpendicular to JK have a slope of +1/2<BR>\n" );
document.write( "To find the equation of the line thru H with a slope of +1/2, use<BR>\n" );
document.write( "y-y1 = m*(x-x1)<BR>\n" );
document.write( "y-2 = (1/2)*(x-5)<BR>\n" );
document.write( "2y-4 = x-5<BR>\n" );
document.write( "x-2y = 1  (eqn of line thru H)<BR>\n" );
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document.write( "Find the eqn of JK:<BR>\n" );
document.write( "Use the same method with either point, J or K<BR>\n" );
document.write( "y+4 = -2*(x+2)<BR>\n" );
document.write( "y = -2x-8<BR>\n" );
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document.write( "Find where the lines intersect by solving the 2 eqns.<BR>\n" );
document.write( "x-2y = 1  (eqn of line thru H)<BR>\n" );
document.write( "y = -2x-8<BR>\n" );
document.write( "Sub for y into eqn thru H<BR>\n" );
document.write( "x - 2(-2x-8) = 1<BR>\n" );
document.write( "x+4x + 16 = 1<BR>\n" );
document.write( "5x = -15<BR>\n" );
document.write( "x = -3<BR>\n" );
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document.write( "y = -2<BR>\n" );
document.write( "--> Intersection at (-3,-2)<BR>\n" );
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document.write( "Find distance from H to intersection:<BR>\n" );
document.write( "d^2 = (5+3)^2 + (2+2)^2 = 64+16<BR>\n" );
document.write( "distance = sqrt(80)<BR>\n" );
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