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Consider three lines L 1 :   5 x - y + 4 = 0 ,   L 2 :   3 x - y + 5 = 0 and L 3 :   x + y + 8 = 0 . If these lines form a triangle A B C , then the sum of the squares of the tangent of all the interior angles of the triangle is

Options

  1. A401 64
  2. B399 64
  3. C111 64
  4. D113 64

Correct answer

A. 401 64

Step-by-step solution

Arranging the lines in descending order of slopes, we get, m 1 = 5 ,   m 2 = 3 and m 3 = - 1 ∴ tan ⁡ A = m 1 - m 2 1 + m 1 m 2 = 2 1 + 15 = 1 8 tan ⁡ B = m 2 - m 3 1 + m 2 m 3 = 3 + 1 1 - 3 = - 2 tan ⁡ C = m 3 - m 1 1 + m 3 m 1 = - 1 - 5 1 - 5 = 3 2 Hence, ∑ t a n 2 A = 1 64 + 4 + 9 4 = 1 + 256 + 144 64 = 401 64

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