document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1168973>Question 1168973</A>: <I><SPAN STYLE=\"word-break: break-all;\"> (7,7) and (9,2)\r<BR>\n" );
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document.write( "I need help finding Leg 1, Leg 2 and the Distance. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #793611 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=793611\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Plot the two points on the same xy grid system. Connect them with a line segment. This line segment will be the hypotenuse of the right triangle we'll set up.\r<BR>\n" );
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document.write( "To get this right triangle, we'll need a third point (where the 90 degree angle is located). There are two possibilities: We could plot a point at (7,2) or at (9,7)\r<BR>\n" );
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document.write( "In this diagram<BR>\n" );
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document.write( "We have A,B as the two given points (7,7) and (9,2) respectively. The other two points C and D are candidates for the third point to pick. \r<BR>\n" );
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document.write( "If you go with point C, then focus on triangle ABC. This is a right triangle with the 90 degree angle at point C.\r<BR>\n" );
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document.write( "The distance from A to C is 5 units because point A has y coordinate 7 and point C has y coordinate 2 (which is also the y coordinate of B), so 7-2 = 5 is the length of AC.\r<BR>\n" );
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document.write( "Similarly, 9-7 = 2 is the distance from C to B. We subtract the x coordinates to get the horizontal distance. \r<BR>\n" );
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document.write( "At this point we know the legs of triangle ABC are 5 and 2. The hypotenuse is unknown but we can use the pythagorean theorem to solve for it.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=25%2B4+=+c%5E2 ALIGN=MIDDLE  ALT=\"25%2B4+=+c%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The distance from point A to B is exactly <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%2829%29 ALIGN=MIDDLE  ALT=\"sqrt%2829%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > units and approximately <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%2829%29+=+5.3851648071345 ALIGN=MIDDLE  ALT=\"sqrt%2829%29+=+5.3851648071345\"   DISABLED_style= \"max-width:90%; height:auto;\" > units.\r<BR>\n" );
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document.write( "Another approach is to use the distance formula. This skips the need for a diagram and plotting that third point.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large (x_1,y_1) = (7,7)\"> is the first point<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large (x_2,y_2) = (9,2)\"> is the second point\r<BR>\n" );
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document.write( "Plug those coordinate values into the distance formula below. Simplify.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d+=+sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2%2B%28y%5B1%5D-y%5B2%5D%29%5E2%29 ALIGN=MIDDLE  ALT=\"d+=+sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2%2B%28y%5B1%5D-y%5B2%5D%29%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d+=+sqrt%28%28-2%29%5E2%2B%285%29%5E2%29 ALIGN=MIDDLE  ALT=\"d+=+sqrt%28%28-2%29%5E2%2B%285%29%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d+=+sqrt%284%2B25%29 ALIGN=MIDDLE  ALT=\"d+=+sqrt%284%2B25%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d+=+sqrt%2829%29 ALIGN=MIDDLE  ALT=\"d+=+sqrt%2829%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > Exact Distance\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d+=+5.3851648071345 ALIGN=MIDDLE  ALT=\"d+=+5.3851648071345\"   DISABLED_style= \"max-width:90%; height:auto;\" > Approximate Distance\r<BR>\n" );
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document.write( "We get the same answer. note the expressions <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B1%5D-x%5B2%5D ALIGN=MIDDLE  ALT=\"x%5B1%5D-x%5B2%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y%5B1%5D-y%5B2%5D ALIGN=MIDDLE  ALT=\"y%5B1%5D-y%5B2%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > compute the length of the legs of the right triangle as done in the previous section (since we subtracted coordinates to get the horizontal and vertical distances). So basically the distance formula is the pythagorean theorem, but a slightly different version. We've gone from <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5E2%2Bb%5E2+=+c%5E2 ALIGN=MIDDLE  ALT=\"a%5E2%2Bb%5E2+=+c%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=c+=+sqrt%28a%5E2%2Bb%5E2%29 ALIGN=MIDDLE  ALT=\"c+=+sqrt%28a%5E2%2Bb%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > where c > 0 is the distance we're after, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a+=+x%5B1%5D-x%5B2%5D ALIGN=MIDDLE  ALT=\"a+=+x%5B1%5D-x%5B2%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b+=+y%5B1%5D-y%5B2%5D ALIGN=MIDDLE  ALT=\"b+=+y%5B1%5D-y%5B2%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
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