SOLUTION: 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^2 -PC^2 =3(PB^2 -QC^2)}}}
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-> SOLUTION: 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^2 -PC^2 =3(PB^2 -QC^2)}}}
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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 Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! On the vertical axis PB = 1/2,PA= 1/2
Let AC= x
1 + x^2 = BC^2
BC = sqrt( 1 + x^2 )
PC = sqrt( .5^2 + x^2 )
BQ = sqrt( 1 + x^2 )/2
QC = sqrt( 1 + x^2 )/2
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BQ^2 - PC^2 = 3*( PB^2 - QC^2 )
( sqrt( 1 + x^2 )/2 )^2 - ( sqrt( .5^2 + x^2 ) )^2 = 3*( .5^2 - ( sqrt( 1 + x^2 )/2 )^2 )
( 1 + x^2 )/4 - 1/4 - x^2 = 3*( 1/4 - ( 1 + x^2 )/4 )
X^2/4 - x^2 = 3/4 *(- x^2 )
-3/4*x^2 = -3/4*x^2
OK
