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Distance Formula

The orthocent...

Question

The orthocentre of the triangle ABC is B and the circumcentre is S (a, b). If A is the origin then the co-ordinates of C are

(2 a, 2 b)

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(\frac{a}{2}, \frac{b}{2})

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(\sqrt{a^{2} + b^{2}}, 0)

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N o n e

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Solution

The correct option is A $(2 a, 2 b)$
$G i v e n -$

$∴ A (0, 0) i s a v e r t e x o f Δ A B C .$

$B i s t h e o r t h o c e n t r e a n d S (a, b) i s t h e c i r c u m c e n t r e o f Δ A B C .$

$T o f i n d o u t - C (x,, y) = ?$

$S o l u t i o n - L e t t h e c o o r d i n a t e s o f C b e C (x, y) . T h e o r t h o c e n t r e i s a t B . i . e Δ A B C i s a r i g h t o n e w i t h$

$∠ B = 90^{0} a n d A C a s h y p o t e n u s e .$

$N o w t h e c i r c u m c e n t r e o f a r i g h t t r i a n g l e i s a t t h e m i d p o i n t o f t h e h y p o t e n u s e . S o, u s i n g m i d p o i n t f o r m u l a,$

$S (a, b) = (\frac{x - 0}{2}, \frac{y - 0}{2})$

$⟹ \frac{x}{2} = a$

$⟹ x = 2 a a n d$

$\frac{y}{2} = b$

$⟹ y = 2 b .$

$∴ C (x, y) = C (2 a, 2 b) . A n s = A .$

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