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  • Vertices of a variable triangle are <img alt="\mathrm{(3,4),(5 \cos \theta, 5 \sin \theta) \: and \: (5 \sin \theta,-5 \cos \theta)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%

Vertices of a variable triangle are \mathrm{(3,4),(5 \cos \theta, 5 \sin \theta) \: and \: (5 \sin \theta,-5 \cos \theta)}, where \mathrm{ \theta \in R}. Locus of it's orthocentre is -
 

Option: 1

\mathrm{(x+y-1)^2+(x-y-7)^2=100}
 


Option: 2

\mathrm{(x+y-7)^2+(x-y-1)^2=100}


 


Option: 3

\mathrm{(x+y-7)^2+(x+y-1)^2=100}
 


Option: 4

\mathrm{(x+y-7)^2+(x-y+1)^2=100}


Answers (1)

best_answer

Distance of all the points from (0,0) are 5 units. That means circumcenter of the triangle formed by the given point is (0,0). If \mathrm{G \equiv(h, k)} be the centroid of triangle, then \mathrm{3 h=3+5 \cos \theta+\sin \theta),}\mathrm{3 \mathrm{k}=4+5(\sin \theta-\cos \theta). If\: H\: (\alpha, \beta)}
be the orthocentre then

\mathrm{\text { OG : GH }=1: 2 \Rightarrow \alpha=3 \mathrm{~h}, \beta=3 \mathrm{k} }

\mathrm{\cos \theta+\sin \theta=\frac{\alpha-3}{5} \quad, \sin \theta-\cos \theta=\frac{\beta-4}{5} }

\mathrm{\Rightarrow \sin \theta=\frac{\alpha+\beta-7}{10}, \cos \theta=\frac{\alpha-\beta+1}{10} }

Thus locus of \mathrm{(\alpha, \beta)\: is \: (x+y-7) 2+(x-y+1) 2=100 }

Hence option 4 is correct.

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mansi

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