The orthocentre of the triangle with vertices 2- -3-1-2- 1-2- -1-2 and 2- -1-2 is

The correct option is B (2, 1/2) #160;Let the Δ be ABC withA=( x _ 1 , y _ 1 ) =( 2,√( 3 ) 1 / 2 #160;) ,B=( x _ 2 , y _ 2 ) ,=( 1 / ...

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Standard XII

Mathematics

Area of Polygon Using Coordinates

The orthocent...

Question

The orthocentre of the triangle with vertices $(2, \frac{\sqrt{3} - 1}{2}), (\frac{1}{2}, - \frac{1}{2})$ and $(2, - \frac{1}{2})$ is

(\frac{3}{2}, \frac{\sqrt{3} - 3}{6})

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(2, - \frac{1}{2})

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(\frac{5}{4}, \frac{\sqrt{3} - 2}{4})

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(\frac{1}{2}, - \frac{1}{2})

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Solution

The correct option is B $(2, - \frac{1}{2})$
$L e t t h e Δ b e A B C w i t h A = (x_{1}, y_{1}) = (2, \frac{\sqrt{3} - 1}{2}), B = (x_{2}, y_{2}), = (\frac{1}{2}, - \frac{1}{2}) a n d C = (x_{3}, y_{3}) = (2, - \frac{1}{2}) . () = \frac{- \frac{1}{2} + \frac{1}{2}}{2 - \frac{1}{2}} = 0.................................. (i i) a n d S l o p e o f A C = m_{A B} = \frac{y_{3} - y_{1}}{x_{3} - x_{1}} = \frac{- \frac{1}{2} - \frac{\sqrt{3} - 1}{2}}{2 - 2} = \infty . . . . . . . . . . . . . . . . . . . . (i i i) F r o m (i i i) A C ⊥ X - a x i s a n d f r o m (i i) B C ∥ X - a x i s ∴ Δ A B C i s a r i g h t t r i a n g l e w i t h C = 1 r i g h t a n g l e w h i c h i s t h e o r t h o c e n t e r s i n c e t h e o r t h o c e n t e r o f a r i g h t t r i a n g l e l i e s a t t h e v e r t e x c o n t a i n i n g t h e r i g h t a n g l e . C o - o r d i n a t e o f t h e o r t h o c e n t e r i s C = (2, - \frac{1}{2}) A n s - O p t i o n C$

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