document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=16749>Question 16749</A>: <I><SPAN STYLE=\"word-break: break-all;\"> what is the distance between\r<BR>\n" );
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document.write( "y=1/2x,y=1/2x+5 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #8465 by </B><A HREF=/tutors/aboutme.mpl?userid=rapaljer><B>rapaljer(4671)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rapaljer><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=rapaljer><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=8465\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> y=1/2x is the graph of a straight line with y intercept 0 and slope m=1/2<BR>\n" );
document.write( "y=1/2x + 5 is the graph of a straight line with y intercept 5 and slope m = 1/2.\r<BR>\n" );
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document.write( "The second line is exactly the same as the first line, except that it is moved \"up\" 5 units.  \r<BR>\n" );
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document.write( "If you rewrite these equations in standard form there is a formula for the distance from a point on a given line to another given line.  Let's rewrite these equations in standard form as <BR>\n" );
document.write( "y = 1/2x,  which is 2y= x  or x-2y=0<BR>\n" );
document.write( "y=1/2x + 5, which is 2y = x + 10 or x-2y +10= 0\r<BR>\n" );
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document.write( "The formula (if I remember correctly from about 15 years ago!!)is : <BR>\n" );
document.write( "Distance from a point to a given line:  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+d+=+%28abs%28Ax%2BBy+%2B+C%29%29%2F%28sqrt%28A%5E2+%2B+B%5E2%29%29+ ALIGN=MIDDLE  ALT=\"+d+=+%28abs%28Ax%2BBy+%2B+C%29%29%2F%28sqrt%28A%5E2+%2B+B%5E2%29%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" >, where Ax+By+C = 0 is the equation of the line, and (x,y) is a given point (in this case, any point you choose to select from the other line--let's say the point (0,0) on the first line, since this is an easy point to see and to calculate).\r<BR>\n" );
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document.write( "So, let (0,0) be a point on the first line, and find the perpendicular distance from (0,0) to the line x-2y+10 = 0.\r<BR>\n" );
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document.write( "In the formula, A= 1, B=-2, C= 10, x=0, and y=0:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+d+=+%28abs%28Ax%2BBy+%2B+C%29%29%2F%28sqrt%28A%5E2+%2B+B%5E2%29%29+ ALIGN=MIDDLE  ALT=\"+d+=+%28abs%28Ax%2BBy+%2B+C%29%29%2F%28sqrt%28A%5E2+%2B+B%5E2%29%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+d+=+%28abs%281%2A0%2B%28-2%29%2A0+%2B+10%29%29%2F%28sqrt%281%5E2+%2B+%28-2%29%5E2%29%29+ ALIGN=MIDDLE  ALT=\"+d+=+%28abs%281%2A0%2B%28-2%29%2A0+%2B+10%29%29%2F%28sqrt%281%5E2+%2B+%28-2%29%5E2%29%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d=+10%2Fsqrt%285%29+=+%2810%2Asqrt%285%29%29%2F%28sqrt%285%29%2Asqrt%285%29%29+ ALIGN=MIDDLE  ALT=\"d=+10%2Fsqrt%285%29+=+%2810%2Asqrt%285%29%29%2F%28sqrt%285%29%2Asqrt%285%29%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d+=+%2810%2Asqrt%285%29%29%2F5 ALIGN=MIDDLE  ALT=\"d+=+%2810%2Asqrt%285%29%29%2F5\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d+=+2%2Asqrt%285%29 ALIGN=MIDDLE  ALT=\"d+=+2%2Asqrt%285%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Someone may have an easier way, but this is what I recall from a LONG time ago!\r<BR>\n" );
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document.write( "R^2 at SCC </SPAN>\n" );
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