Q.
Vertices of a variable triangle are (3 , 4) (5cosθ,5sinθ) and (5sinθ,−5cosθ) . Then locus of its orthocenter is
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a
(x+y−1)2+(x−y−7)2=100
b
(x+y−7)2+(x−y−1)2=100
c
(x+y−7)2+(x+y−1)2=100
d
(x+y−7)2+(x−y+1)2=100
answer is D.
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Detailed Solution
Circumcentre of the triangle is (0 , 0) and Centroid (3+5cosθ+5sinθ3,4+5sinθ−5cosθ3) [∵ Centroid divides the join of orthocenter and circumentre in 2 : 1 ratio ⇒ G=2S+O3 If S=(0,0) then 3G=Olet (h , k) represents orthocenter and 3G=3+5cosθ+5sinθ ,4+5sinθ-5cosθ ∴h=3+5cosθ+5sinθ k=4+5sinθ−5cosθ Adding and subtracting the above equations we get ⇒sinθ=h+k−710;cosθ=h−k+110 squarring and adding ⇒(h+k−7)2+(h−k+1)2=100 ∴ Locus of orthocenter is (x+y−7)2+(x−y+1)2=100
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