document.write( "Question 144172: Given: triangle ABC A(-1,2) B(7,0) C(1,-6) and a point D(4,-3) on segment BC
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Algebra.Com's Answer #105184 by Edwin McCravy(20054)\"\" \"About 
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Given: triangle ABC A(-1,2) B(7,0) C(1,-6) and a point D(4,-3) on segment BC
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document.write( "First we use the midpoint formula to show that D \r\n" );
document.write( "is the midpoint of BC.  That will show that AD is \r\n" );
document.write( "a bisector of BC.  Then we will use the slope \r\n" );
document.write( "formula to show that AD is perpendicular to BC.\r\n" );
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document.write( "The midpoint of the segment joining (\"x1\",\"y1\") and (\"x2\",\"y2\") is\r\n" );
document.write( "given by the formula\r\n" );
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document.write( "\"MIDPOINT\" = (\"%28x1%2Bx2%29%2F2\",\"%28y1-y2%29%2F2\")\r\n" );
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document.write( "We use B(\"7\",\"0\") as (\"x1\",\"y1\")\r\n" );
document.write( "and C(\"1\",\"-6\") as (\"x2\",\"y2\") \r\n" );
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document.write( "\"MIDPOINT\" = (\"%287%2B1%29%2F2\",\"%280%2B%28-6%29%29%2F2\")\r\n" );
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document.write( "\"MIDPOINT\" = (\"8%2F2\",\"%280-6%29%2F2\")\r\n" );
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document.write( "\"MIDPOINT\" = (\"4\",\"-6%2F2\")  \r\n" );
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document.write( "\"MIDPOINT\" = (\"4\",\"-3\")\r\n" );
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document.write( "Since D has those coordinates, AD bisects BC.\r\n" );
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document.write( "Now we need to show AD and BC are perpendicular.\r\n" );
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document.write( "The slope of the segment joining (\"x1\",\"y1\") and (\"x2\",\"y2\") is\r\n" );
document.write( "given by the formula\r\n" );
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document.write( "\"m+=+%28y2-y1%29%2F%28x2-x1%29\"\r\n" );
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document.write( "So we now find the slope of BC, again using\r\n" );
document.write( "B(\"7\",\"0\") as (\"x1\",\"y1\")\r\n" );
document.write( "and C(\"1\",\"-6\") as (\"x2\",\"y2\")\r\n" );
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document.write( "\"m+=+%28y2-y1%29%2F%28x2-x1%29\"\r\n" );
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document.write( "\"m+=+%28%28-6%29-%280%29%29%2F%28%281%29-%287%29%29\"\r\n" );
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document.write( "\"+m+=+%28-6%29%2F%28-6%29\"\r\n" );
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document.write( "\"m+=+1\"\r\n" );
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document.write( "So the slope of BC is \"1\"\r\n" );
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document.write( "Now So we now find the slope of AD, using\r\n" );
document.write( "A(\"-1\",\"2\") as (\"x1\",\"y1\")\r\n" );
document.write( "and D(\"4\",\"-3\") as (\"x2\",\"y2\")\r\n" );
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document.write( "\"m+=+%28%28-3%29-%282%29%29%2F%28%284%29-%28-1%29%29\"\r\n" );
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document.write( "\"m+=+%28-3-2%29%2F%28%284%29%2B%281%29%29\"\r\n" );
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document.write( "\"+m+=+%28-5%29%2F5\"\r\n" );
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document.write( "\"m+=+-1\"\r\n" );
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document.write( "So the slope of AD is \"-1\"\r\n" );
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document.write( "Since \"1\" and \"-1\" are reciprocals with \r\n" );
document.write( "opposite signs, this proves AD is perpendicular to\r\n" );
document.write( "BC.\r\n" );
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document.write( "Therefore AD is the perpendicular bisector of BC.\r\n" );
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document.write( "Edwin
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