Let A ≡ (3, 2) and B ≡ (5, 1). ABP is an equilateral triangle isconstructed on the side of AB remote from the origin then the orthocentre of triangle ABP is - (A) 4 - 1 2 3 , 3 2 - 3 (B) 4 + 1 2 3 , 3 2 + 3 (C) 4 - 1 6 3 , 3 2 - 1 3 3 (D) 4 + 1 6 3 , 3 2 + 1 3 3
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The answer to this question is D
Mab = -1/2
mpm = 2
h = a√3/2
now a = √5 = AB; h√15/2
so point P x-y/cosθ = y -3/2/sinθ = h
P: x = 4 + √3 , y = 3/2 + √3
so orthocentre/centroid is ∑x1/3 , ∑y1/3
= 4 + √3/6 , 3/2 + √3/2
Mab = -1/2
mpm = 2
h = a√3/2
now a = √5 = AB; h√15/2
so point P x-y/cosθ = y -3/2/sinθ = h
P: x = 4 + √3 , y = 3/2 + √3
so orthocentre/centroid is ∑x1/3 , ∑y1/3
= 4 + √3/6 , 3/2 + √3/2
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