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) ≠ 0, then (x, y) lies on the straight line whose equation is:
(A) 3x – 4y – 1 = 0 (B) 4x – 3y – 4 = 0
(C) 4x – 3y – 1 = 0 (D) 3x – 4y – 4 = 0

Answer 1

Answer:

Correct option (1) 4x – 3y – 4 = 0

Explanation:

x – 2y + kz = 1 …. (i) 2x + y + z = 2 …

(ii) 3x – y – kz = 3 …

(iii) for locus of (x, y) equation (i) + (iii) 4x – 3y = 4

4x – 3y – 4 = 0.

Answer 2

Answer:

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Correct option (B)✔