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JEE Main20198 Apr 2019Evening ShiftMathematicsDeterminantsActual

If the system of linear equations x - 2 y + k z = 1 2 x + y + z = 2 3 x - y - k z = 3 has a solution x , y , z , z ≠ 0 , then x , y lies on the straight line whose equation is:

Options

  1. A4 x - 3 y - 4 = 0
  2. B3 x - 4 y - 4 = 0
  3. C3 x - 4 y - 1 = 0
  4. D4 x - 3 y - 1 = 0

Correct answer

A. 4 x - 3 y - 4 = 0

Step-by-step solution

x - 2 y + k z = 1 .........(1) 2 x + y + z = 2 .........(2) 3 x - y - k z = 3 ........(3) Hence, 1 - 2 k 2 1 1 3 - 1 - k x y z = 1 2 3 For infinite solution, 1 - 2 k 2 1 1 3 - 1 - k = 0 ⇒ - k + 1 + 2 - 2 k - 3 + k - 2 - 3 = 0 ⇒ k = - 1 2 Put this value in the equation 1 , we get 2 x - 4 y - z = 2 . . . . 4 Adding the equation 4   &   2 , we get 4 x - 3 y = 4 Hence, the equation of straight line is 4 x - 3 y - 4 = 0

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