Find the equations of any two altitudes of the triangle whose angular points are A(7,-1), B(-2,8), C(1,2). Hence find the orthocentre of the triangle

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Solution

$S l o p e o f l i n e B C = (2 - 8) / (1 + 2) = - 6 / 3 = - 2 N o w s l o p e o f a n y p e r p e n d i c u l a r t o l i n e B C i s g i v e n b y - 1 / (- 2) = 1 / 2 N o w, e q u a t i o n o f a l i n e p e r p e n d i c u l a r t o B C a n d p a s s i n g t h r o u g h A (7, - 1) a n d o f s l o p e 1 / 2 i s g i v e n b y (y + 1) / (x - 7) = 1 / 2 \Rightarrow 2 (y + 1) = (x - 7) \Rightarrow 2 y - x + 2 + 7 = 0 \Rightarrow 2 y - x + 9 = 0 - - - - - (1) S i m i l a r l y s l o p e o f l i n e A B = (8 + 1) / (- 2 - 7) = 9 / (- 9) = - 1 t h u s, s l o p e o f l i n e p e r p e n d i c u l a r t o l i n e A B i s g i v e n b y = - 1 / (- 1) = 1 N o w e q u a t i o n o f l i n e p a s s i n g t h r o u g h C (1, 2) a n d p e r p e n d i c u l a r t o A B i s g i v e n b y (y - 2) / (x - 1) = 1 \Rightarrow y - 2 = x - 1 \Rightarrow y - x - 1 = 0 - - - (2) N o w, o r t h o c e n t r e i s g i v e n b y t h e p o i n t o f i n t e r s e c t i o n o f e q (1) a n d e q (2) S u b t r a c t i n g e q (2) f r o m e q (1) w e g e t y + 9 + 1 = 0 \Rightarrow y = - 10 y - x - 1 = 0 \Rightarrow x = y - 1 = - 10 - 1 = - 11 T h u s o r t h o c e n t r e i s (- 11, - 10)$

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