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) \[{{(x+y-7)}^{2}}+{{(x-y+1)}^{2}}=100\]
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 \[{{(x+y-7)}^{2}}+{{(x-y+1)}^{2}}=100.\]
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