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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Vertices of a variable triangle are (3 , 4) (5cosθ,5sinθ)  and (5sinθ,−5cosθ) . Then locus of its orthocenter is