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Let A 6,7 , B 2,3 and C - 2,1 be the vertices of a triangle. The point P nearer to the point A such that Δ P B C is an equilateral triangle is

Options

  1. A- 3 , 2 + 2 3
  2. B3 , 2 + 2 3
  3. C3 , 2 - 2 3
  4. D- 3 , 2 - 2 3

Correct answer

A. - 3 , 2 + 2 3

Step-by-step solution

We have, B C = 2 5 . Since, Δ P B C is an equilateral triangle and the point P lies in the interior of Δ A B C . Therefore, P lies on the perpendicular bisector of B C at a distance of 3 2 B C = 15 from the mid-point of B C such that P and A are on the same side of B C . The equation of B C is x - 2 y + 4 = 0 The coordinates of the mid-point of B C are D 0,2 The slope of a line perpendicular to B C is - 2 . So, if it makes an angle θ with the x -axis. Then, tan ⁡ θ = - 2 ⇒ sin ⁡ θ = 2 5 and cos ⁡ θ = - 1 5 The para

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