Q.422. To prove an identity related to triangle.

Question 422.
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).

Answer 422.
Let origin be at A and x-axis be along AB and C on y-axis
=> A = (0, 0), B = (2a, 0) and C = (0, 2b)
=> P = (a, 0) and Q = (a, b)
=> BQ^2 - PC^2
= (2a - a)^2 + (0- b)^2 - [(a - 0)^2 + (0 - 2b)^2]
= - 3b^2 ... ( 1 )
and 3(PB^2 - QC^2)
= 3 [(2a - a)^2 + 0^2 - (a - 0)^2 - (b - 2b)^2]
= - 3b^2 ... ( 2 )
From ( 1 ) and ( 2 ),
BQ^2 - PC^2 = 3(PB^2 - QC^2).

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