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One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is

(a) (–1, –1)

(b) (2, 2)

(c)  (–2, –2)

(d) (2, –2)

0 3 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    3 months ago

    This is a Objective Type Questions as classified in NCERT Exemplar

    Sol:

    L e t     A B C     b e     a n     e q u i l a t e r a l     t r i a n g l e     w i t h     v e r t e x     ( x 1 , y 1 ) . A D ⊥ B C     a n d     l e t     ( a , b )     b e     t h e     c o o r d i n a t e s     o f     D . G i v e n     t h a t     t h e     c e n t r o i d     G     l i e s     a t     t h e     o r i g i n     i . e . , ( 0 , 0 ) Since,  the  centroid  of  a  triangle,  divides  the  median  in  the  ratio  1:2 S o ,                         0 = 1 × x 1 + 2 × a 1 + 2                     ⇒ x 1 + 2 a = 0                                                                                           … ( i ) a n d                       0 = 1 × y 1 + 2 × b 1 + 2                   ⇒ y 1 + 2 b = 0                                                                                           … ( i i ) E q u a t i o n s     o f     B C     i s     g i v e n     b y     x + y − 2 = 0                                                                                       … ( i i i ) Point  D(a,b)  lies  on  the  line  x+y−2=0 S o ,                                                                                                                     a + b − 2 = 0                                                                                           … ( i v ) S l o p e     o f     e q n . ( i i i )     i s     = − 1 a n d     t h e     S l o p e     o f     A G = y 1 − 0 x 1 − 0 = y 1 x 1 Since,  they  are  perpendicular  to  each  other ∴                       − 1 × y 1 x 1 = − 1       ⇒ y 1 = x 1 F r o m     e q n . ( i )     a n d     ( i i )     w e     g e t                               x 1 + 2 a = 0         ⇒ 2 a = − x 1                               y 1 + 2 b = 0         ⇒ 2 b = − y 1 F r o m     e q n . ( i v )     w e     g e t                         a + b − 2 = 0 ⇒               a + a − 2 = 0 ⇒                         2 a − 2 = 0             ⇒ a = 1     a n d     b = 1               [ ? a = b ] ∴               x 1 = − 2 × 1 = − 2 a n d     y 1 = − 2 × 1 = − 2 H e n c e ,     t h e     c o r r e c t     o p t i o n     i s     ( c ) .

     

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