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Vertices of a variable triangle are $\mathrm{(5,12),(-13 \cos \theta, 13 \sin \theta), and \: (13 \sin \theta,-13 \cos \theta)}$ . Then locus of its orthocentre is

Option: 1
$\mathrm{x-y+7=0}$

Option: 2
$\mathrm{x-y-7=0}$

Option: 3
$\mathrm{x+y-7=0}$

Option: 4
$\mathrm{x+y+7=0}$

Answers (1)

best_answer

Circumcentre is $(0,0)$

$\mathrm{ \text { Centroid G is }\left(\frac{5-13 \cos \theta+13 \sin \theta}{3}, \frac{12+13 \sin \theta-13 \cos \theta}{3}\right) }$

$\mathrm{ \text { G divides HO internally } 2: 1 }$

$\mathrm{ \frac{5-13 \cos \theta+13 \sin \theta}{3}=\frac{2.0+1 . h}{3} }$

$\mathrm{ \frac{12+13 \sin \theta-13 \cos \theta}{3}=\frac{2.0+1 . k}{3} }$
$\mathrm{ \Rightarrow \quad \sin \theta-\cos \theta=\frac{h-5}{13} }$

$\mathrm{ \sin \theta-\cos \theta=\frac{k-12}{13} }$

$\mathrm{ \Rightarrow \text { from (1) and (2), } \frac{h-5}{13}=\frac{k-12}{13} }$

$\mathrm{ \Rightarrow \text { h } k+7=0 }$

Locus of orthocentre $\mathrm{ H(h, k)\: is }$

$\mathrm{ \: x-y+7=0 }$

Hence option 1 is correct.

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