document.write( "Question 1120904: Triangle ABC is right-angled at a. P is the midpoint of AB and Q is the midpoint of BC. Choose suitable coordinates in order to prove that $\"BQ%5E2+-PC%5E2+=3%28PB%5E2+-QC%5E2%29\"$ \n" ); document.write( "

Algebra.Com's Answer #736596 by josmiceli(19441) $\"\"$ $\"About$

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On the vertical axis PB = 1/2,PA= 1/2
\n" ); document.write( "Let AC= x
\n" ); document.write( "1 + x^2 = BC^2
\n" ); document.write( "BC = sqrt( 1 + x^2 )
\n" ); document.write( "PC = sqrt( .5^2 + x^2 )
\n" ); document.write( "BQ = sqrt( 1 + x^2 )/2
\n" ); document.write( "QC = sqrt( 1 + x^2 )/2
\n" ); document.write( "覧覧覧覧覧覧覧-
\n" ); document.write( "BQ^2 - PC^2 = 3*( PB^2 - QC^2 )
\n" ); document.write( "( sqrt( 1 + x^2 )/2 )^2 - ( sqrt( .5^2 + x^2 ) )^2 = 3*( .5^2 - ( sqrt( 1 + x^2 )/2 )^2 )
\n" ); document.write( "( 1 + x^2 )/4 - 1/4 - x^2 = 3*( 1/4 - ( 1 + x^2 )/4 )
\n" ); document.write( "X^2/4 - x^2 = 3/4 *(- x^2 )
\n" ); document.write( "-3/4*x^2 = -3/4*x^2
\n" ); document.write( "OK
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