If the equations 4x + (k − 2)y + 3 = 0 and (k − 2)x + 9y − 5 = 0 (k > 0) are parallel, then find the value of k 2 + 6.

Correct — C. Rule: a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel when a₁/a₂ = b₁/b₂.

4/(k − 2) = (k − 2)/9
Cross-multiply: (k − 2)² = 36
k − 2 = ±6 → k = 8 or k = −4
The stem sets k > 0, so k = 8

The question does not ask for k:
k² + 6 = 64 + 6 = 70 → option (c)

Why the others are wrong

• (a)
  72 — 72 is 64 + 8. The square is right and the constant added is k itself rather than the 6 the question names.
• (b)
  68 — 68 is 64 + 4. Right square, wrong constant — 4 is the coefficient of x in the first equation and has no part in k² + 6.
• (d)
  64 — 64 is k², with the + 6 never added. Solving for k is the hard part of the question, which is exactly why the unfinished answer is on the list.

Two lines written as ax + by + c = 0 are parallel when their x- and y-coefficients share a ratio, a₁/a₂ = b₁/b₂, and are distinct provided c₁/c₂ differs from it. Equal slopes, different intercepts.

Here a₁/a₂ = 4/(k − 2) and b₁/b₂ = (k − 2)/9, so the condition is a quadratic in k, not a linear equation. That is what puts two roots on the table.

Both roots give genuinely parallel lines; the stated k > 0 is the only thing selecting k = 8 over k = −4.

Key facts

• Parallel lines satisfy a₁/a₂ = b₁/b₂ while c₁/c₂ differs from that ratio.
• If a₁/a₂ = b₁/b₂ = c₁/c₂ the two equations describe one and the same line.
• Here (k − 2)² = 36 gives k = 8 or k = −4, and the stated k > 0 keeps only k = 8.
• k = 8 makes the lines 4x + 6y + 3 = 0 and 6x + 9y − 5 = 0.

Common traps

• Answering 64 — finding k and forgetting the + 6 the question actually asks for.
• Setting coefficients equal instead of proportional: 4 = k − 2 gives k = 6, which is not parallel.
• Reaching for the perpendicular condition a₁a₂ + b₁b₂ = 0 by reflex.

No directly related past PYQ was found.