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Question Number 56390 by Tinkutara last updated on 15/Mar/19

Commented by Tinkutara last updated on 18/Mar/19
Answer is
Commented by Tinkutara last updated on 18/Mar/19

	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19
	$A(1,2) B((√(5 )) cosθ,(√5) sinθ) C((√5) sinθ,−(√5) cosθ) slope of AB →m_1 =(((√5) sinθ−2)/((√5) cosθ−1)) slope BC→m_2 =(((√5) sinθ+(√5) cosθ)/((√5) cosθ−(√5) sinθ))=((1+tanθ)/(1−tanθ))=tan((π/4)+θ) slope of CA→m_3 =((2+(√5) cosθ)/(1−(√5) sinθ)) eqn st line ⊥BC and passing through eqn AD⊥BC (y−2)=(((−1)/m_2 ))(x−1) (y−2)=(((tanθ−1)/(tanθ+1)))(x−1) eqn BE⊥AC (y−(√5) sinθ)=−((1/m_3 ))(x−(√5) cosθ) (y−(√5) sinθ)=((((√5) sinθ−1)/(2+(√5) cosθ)))(x−(√5) cosθ) solving eqn AD and BE we get ortho centre 2+(((tanθ−1)/(tanθ+1)))(x−1)=(√5) sinθ+((((√5) sinθ−1)/(2+(√5) cosθ)))(x−(√5) cosθ) for easy of simplificatiin (√5) sinθ=a (√5) cosθ=b tanθ=(a/b) 2+((((a/b)−1)/((a/b)+1)))(x−1)=a+(((a−1)/(2+b)))(x−b) 2+x(((a−b)/(a+b)))−(((a−b)/(a+b)))=a+x(((a−1)/(2+b)))−b(((a−1)/(2+b))) x{(((a−b)/(a+b)))−(((a−1)/(2+b)))}=a−b(((a−1)/(2+b)))−2+((a−b)/(a+b)) x{((2a−2b+ab−b^2 −a^2 −ab+a+b)/((a+b)(2+b))}=((a(2a+2b+ab+b^2 )−b(a^2 +ab−a−b)−2(2a+2b+ab+b^2 )+(2a+ab−2b−b^2 ))/((2+b)(a+b))) x(3a−b−b^2 −a^2 )=4a+4b+a^2 b+ab^2 −a^2 b−ab^2 +ab+b^2 −4a−4b−2ab−b^2 +2a+ab−2b−b^2 x(3a−b−b^2 −a^2 )=2a−2b+ab+b^2 −2ab−b^2 +ab−b^2 x(3a−b−b^2 −a^2 )=2a−2b−b^2 x=((2a−2b−b^2 )/(3a−b−a^2 −b^2 )) [a=(√5) sinθ b=(√5) cosθ] pls check upto this lengthy question...$
	$A (1, 2) B (\sqrt{5} c o s θ, \sqrt{5} s i n θ) C (\sqrt{5} s i n θ, - \sqrt{5} c o s θ)$ $s l o p e o f A B \to m_{1} = \frac{\sqrt{5} s i n θ - 2}{\sqrt{5} c o s θ - 1}$ $s l o p e B C \to m_{2} = \frac{\sqrt{5} s i n θ + \sqrt{5} c o s θ}{\sqrt{5} c o s θ - \sqrt{5} s i n θ} = \frac{1 + t a n θ}{1 - t a n θ} = t a n (\frac{π}{4} + θ)$ $s l o p e o f C A \to m_{3} = \frac{2 + \sqrt{5} c o s θ}{1 - \sqrt{5} s i n θ}$ $e q n s t l i n e ⊥ B C a n d p a s s i n g t h r o u g h$ $e q n A D ⊥ B C (y - 2) = (\frac{- 1}{m_{2}}) (x - 1)$ $(y - 2) = (\frac{t a n θ - 1}{t a n θ + 1}) (x - 1)$ $e q n B E ⊥ A C$ $(y - \sqrt{5} s i n θ) = - (\frac{1}{m_{3}}) (x - \sqrt{5} c o s θ)$ $(y - \sqrt{5} s i n θ) = (\frac{\sqrt{5} s i n θ - 1}{2 + \sqrt{5} c o s θ}) (x - \sqrt{5} c o s θ)$ $s o l v i n g e q n A D a n d B E w e g e t o r t h o c e n t r e$ $2 + (\frac{t a n θ - 1}{t a n θ + 1}) (x - 1) = \sqrt{5} s i n θ + (\frac{\sqrt{5} s i n θ - 1}{2 + \sqrt{5} c o s θ}) (x - \sqrt{5} c o s θ)$ $f o r e a s y o f s i m p l i f i c a t i i n \sqrt{5} s i n θ = a$ $\sqrt{5} c o s θ = b t a n θ = \frac{a}{b}$ $2 + (\frac{\frac{a}{b} - 1}{\frac{a}{b} + 1}) (x - 1) = a + (\frac{a - 1}{2 + b}) (x - b)$ $2 + x (\frac{a - b}{a + b}) - (\frac{a - b}{a + b}) = a + x (\frac{a - 1}{2 + b}) - b (\frac{a - 1}{2 + b})$ $x {(\frac{a - b}{a + b}) - (\frac{a - 1}{2 + b})} = a - b (\frac{a - 1}{2 + b}) - 2 + \frac{a - b}{a + b}$ $x {\frac{2 a - 2 b + a b - b^{2} - a^{2} - a b + a + b}{(a + b) (2 + b}} = \frac{a (2 a + 2 b + a b + b^{2}) - b (a^{2} + a b - a - b) - 2 (2 a + 2 b + a b + b^{2}) + (2 a + a b - 2 b - b^{2})}{(2 + b) (a + b)}$ $x (3 a - b - b^{2} - a^{2}) = 4 a + 4 b + a^{2} b + {a b}^{2} - a^{2} b - {a b}^{2} + a b + b^{2} - 4 a - 4 b - 2 a b - b^{2} + 2 a + a b - 2 b - b^{2}$ $x (3 a - b - b^{2} - a^{2}) = 2 a - 2 b + a b + b^{2} - 2 a b - b^{2} + a b - b^{2}$ $x (3 a - b - b^{2} - a^{2}) = 2 a - 2 b - b^{2}$ $x = \frac{2 a - 2 b - b^{2}}{3 a - b - a^{2} - b^{2}} [a = \sqrt{5} s i n θ b = \sqrt{5} c o s θ]$ $p l s c h e c k u p t o t h i s l e n g t h y q u e s t i o n . . .$
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