Let ABC be triangle having orthocenter, circumcenter at $(9, 5) (0, 0)$ respectively if equation of side $B C$ is $2 x - y = 10$ then find possible coordinate of vertex $A$ .

(-3, -3)

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(5, 5)

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(4, 4)

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(4,15)

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Solution

The correct option is D (4,15)
$\begin{matrix} L e t t h e c o o r d i n a t e s o f v e r t i c e s A, B a n d C b e (x_{1}, y_{1}), (x_{2}, y_{2}) a n d (x_{3}, y_{3}) A s w e k n o w t h a t c e n t r o i d G, d i v i d e s o r t h o c e n t r e O a n d c i r c u m c e n t r e C i n t h e r a t i o 2 : 1 s o, c o o r d i n a t e s o f G = (\frac{2 \times 0 + 9 \times 1}{2 + 1}, \frac{2 \times 0 + 5 \times 1}{2 + 1}) = (3, \frac{3}{5}) N o w, l e t t h e p e r p e n d i c u l a r b i s e c t o r o f l i n e B C b e D h a v i n g c o o r d i n a t e s (x, y) t h e n y = 2 x - 10 s o, D = (x, 2 x - 10) A l s o, C^{'} D a n d B C a r e p e r p e n d i c u l a r ∴ s l o p e o f C^{'} D \times s l o p e o f B C = - 1 F r o m s o l v i n g w e g e t x = \frac{5}{2} s o D = (\frac{5}{2}, - 5) N o w, \frac{x_{2} + x_{3}}{2} = \frac{5}{2} a n d, \frac{y_{2} + y_{3}}{2} = - 5 x_{2} + x_{3} = 5 a n d y_{2} + y_{3} = 10 - - - - - (1) N o w, C e n t r o i d G = (3, \frac{5}{2}) = [(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})] x_{1} + x_{2} + x_{3} = 9 a n d y_{1} + y_{2} + y_{3} = 5 - - (2) F r o m s o l v i n g e q u a t i o n (1) a n d (2) w e g e t x_{1} = 4 a n d y_{1} = 15 H e n c e, A = (4, 15) \end{matrix}$
Hence, the option $D$ is the correct answer.

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