If the system of linear equations 

x-2y+kz=1

2x+y+z=2

3x-y-kz=3

has a solution (z,y,z), z\neq 0 then (x,y) lies on the straight line whose equation is :

  • Option 1)

    3x-4y-1=0

  • Option 2)

    4x-3y-4=0

  • Option 3)

    4x-3y-1=0

  • Option 4)

    3x-4y-4=0

 

Answers (1)

x-2y+kz=1............(1)

2x+y+z=2............(2)

3x-y-kz=3............(3)

add (1) and (3)

4x-3y=4

 


Option 1)

3x-4y-1=0

Option 2)

4x-3y-4=0

Option 3)

4x-3y-1=0

Option 4)

3x-4y-4=0

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