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  • Vertices of a triangle are <img alt="(5,12),(13 \sin \theta, 13 \cos \theta)" src="https://entrancecorner.oncodecogs.com/gif.latex?%285%2C12%29%2C%2813%20%5Csin%20%5Ctheta%2C%2013%20%5Ccos%20%5Cthe

Vertices of a triangle are (5,12),(13 \sin \theta, 13 \cos \theta) and (13 \cos \theta, 13 \sin \theta), where \theta \in R,then locus of it's orthocentre is

Option: 1

\mathrm{x+y+7=0}


Option: 2

\mathrm{x-y+7=0}


Option: 3

\mathrm{(x+y-1)^{2}+(x-y+5)^{2}=26}


Option: 4

\mathrm{(x-y+5)^{2}+(x+y+7)^{2}=100}


Answers (1)

best_answer

Vertices are at equal distance from origin, so circumcentre of triangle is origin and centroid of triangle is

\mathrm{\left(\frac{5+13 \sin \theta+13 \cos \theta}{3}, \frac{12+13 \sin \theta+13 \cos \theta}{3}\right)}


Centroid divides orthocentre and circumcentre in the ratio 2:1
\mathrm{\Rightarrow \quad \frac{5+13 \sin \theta+13 \cos \theta}{3}=\frac{\bar{x}}{3} ; \frac{12+13 \sin \theta+13 \cos \theta}{3}=\frac{\bar{y}}{3}}
\mathrm{\Rightarrow \quad 13 \sin \theta+13 \cos \theta=\bar{x}-5 ; 13 \sin \theta+13 \cos \theta=\bar{y}-12}
\mathrm{\Rightarrow \quad \bar{x}-5=\bar{y}-12 \Rightarrow x-y+7=0}

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Rishi

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