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Question:

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

4x−3y−1;=0

4x−3y−4=0

3x−4y−1;=0

3x−4y−4=0

Solution:

Correct option is C.
4x−3y−4=0
x - 2y + kz = 1.. (1)
2x + y + z = 2.. (2)
3x - y - kz = 3.. (3)
(1) + (3) ⇒ 4x - 3y = 4