document.write( "Question 698546: if p is the length if the perpendicular from the origin to the line x/a + y/b = 1, prove that 1/p² = 1/a² + 1/b².\r
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Algebra.Com's Answer #430983 by Edwin McCravy(20063)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "The formula for the perpendicular distance from (x1,y1)\r\n" );
document.write( "to the line whose equation is Ax+By+C=0 is\r\n" );
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document.write( "d = \"abs%28%28Ax%5B1%5D%2BBy%5B1%5D%2BC%29%2Fsqrt%28A%5E2%2BB%5E2%29%29\"\r\n" );
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document.write( "The line has the equation\r\n" );
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document.write( "\"x%2Fa\" + \"y%2Fb\" = 1  \r\n" );
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document.write( "[Don't get capital A and B confused with little letters a and b]\r\n" );
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document.write( "Clear of fractions by multiplying by LCD = ab:\r\n" );
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document.write( "     bx + ay = ab\r\n" );
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document.write( "Get 0 on the right\r\n" );
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document.write( "bx + ay - ab = 0\r\n" );
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document.write( "For (x1,y1) = (0,0)\r\n" );
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document.write( "Substitute in\r\n" );
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document.write( "d = \"abs%28%28Ax%5B1%5D%2BBy%5B1%5D%2BC%29%2Fsqrt%28A%5E2%2BB%5E2%29%29\"\r\n" );
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document.write( "with d = p, A = b, B = a, C = -ab, (x1,y1) = (0,0)\r\n" );
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document.write( "p = \"abs%28%28b%2A%280%29%2Ba%2A%280%29-ba%29%2Fsqrt%28b%5E2%2Ba%5E2%29%29\"\r\n" );
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document.write( "p = \"abs%28-ba%2Fsqrt%28b%5E2%2Ba%5E2%29%29\"\r\n" );
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document.write( "Square both sides\r\n" );
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document.write( "p² = \"b%5E2a%5E2%2F%28b%5E2%2Ba%5E2%29\"\r\n" );
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document.write( "Take the reciprocal of both sides:\r\n" );
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document.write( "\"1%2Fp%5E2\" = \"%28b%5E2%2Ba%5E2%29%2F%28b%5E2a%5E2%29\"\r\n" );
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document.write( "Make two fractions on the right:\r\n" );
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document.write( "\"1%2Fp%5E2\" = \"b%5E2%2F%28b%5E2a%5E2%29%2Ba%5E2%2F%28b%5E2a%5E2%29\"\r\n" );
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document.write( "\"1%2Fp%5E2\" = \r\n" );
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document.write( "\"1%2Fp%5E2\" = \"1%2Fa%5E2%2B1%2Fb%5E2\"\r\n" );
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document.write( "Edwin
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