If (0, 0), (3, 0) and (x, y) are the vertices of an equilateral triangle, then the value of x and y is

(\frac{3}{2}, \frac{3 \sqrt{3}}{2}) o r (_{2}^{3}, \frac{- 3 \sqrt{3}}{2})

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(\frac{1}{2}, \sqrt{2}) o r (\frac{1}{2}, - \sqrt{2})

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(\frac{1}{3}, 1) o r (\frac{1}{3}, - 1)

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none of these

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Solution

The correct option is A $(\frac{3}{2}, \frac{3 \sqrt{3}}{2}) o r (_{2}^{3}, \frac{- 3 \sqrt{3}}{2})$
$L e t u s t a k e t h e e q u i l a t e r a l Δ a s A O B w i t h A = (x, y), B = (3, 0), O = (0, 0) . T h e o r d i n a t e s o f B & O a r e = 0. S o B O l i e s o n t h e X - a x i s . A l s o (A B = B O & A O = B O) . . . . . . . . . . . . . . . . . (i) [s i n c e Δ A O B i s e q u i l a t e r a l] ∴ T h e l e n g t h B O = 3 - 0 = 3......... (i i) N o w A B = \sqrt{{(3 - x)}^{2} + {(0 - y)}^{2}} = \sqrt{x^{2} + y^{2} - 6 x + 9} = 3 [f r o m (i)] ⟹ x^{2} + y^{2} - 6 x = 0............. (i i i) A g a i n A O = \sqrt{{(0 - x)}^{2} + {(0 - y)}^{2}} = 3 [f r o m (i)] ⟹ x^{2} + y^{2} = 9................ (i v) ∴ F r o m (i) & (i v) w e h a v e 9 - 6 x = 0 ⟹ x = \frac{3}{2} S u b s t i t u t i n g x = \frac{3}{2} i n (i v), y = \pm \frac{3 \sqrt{3}}{2} ∴ (x, y) = (\frac{3}{2}, \frac{3 \sqrt{3}}{2}), (\frac{3}{2}, - \frac{3 \sqrt{3}}{2}) A n s - O p t i o n A$

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