Straight Line Question 240

Question: Vertices of a variable triangle are $ (3,4), $ $ (5cos\theta ,5sin\theta ) $ and $ (5sin\theta ,-5cos\theta ), $ where $ \theta \in R. $ Locus of its orthocenter is

Options:

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 $

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Answer:

Correct Answer: D

Solution:

  • [d] Distance of all the points from (0, 0) are 5 unit. That means circumventer of the triangle formed by the given point is (0, 0). If G(h, k) be the centroid of triangle, then $ 3h=3+5(cos\theta +sin\theta ),3k=4+5(sin\theta -cos\theta ) $ If $ H(\alpha ,\beta ) $ be the orthocenter, then OG: GH $ =1:2\Rightarrow \alpha =3h,\beta =3k $ $ \cos \theta +sin\theta =\frac{\alpha -3}{5},\sin \theta -\cos \theta =\frac{\beta -4}{5} $

$ \Rightarrow \sin \theta =\frac{\alpha +\beta -7}{10},\cos \theta =\frac{\alpha -\beta +1}{10} $ Thus, locus of $ (\alpha ,\beta ) $ is $ {{(x+y-7)}^{2}}+{{(x-y+1)}^{2}}=100. $