The relation between K₁ and K₂ for which the system of linear equations K₁x + 3y = 8 and 4x + K₂y = 16 represents coincident lines, is:

Correct — A. Coincident lines need all three ratios equal: K₁⁄4 = 3⁄K₂ = 8⁄16.

Constant ratio: 8⁄16 = 1⁄2

K₁⁄4 = 1⁄2, so K₁ = 2
3⁄K₂ = 1⁄2, so K₂ = 6

Test the relation: 6 = 3 × 2, so K₂ = 3K₁ → option (a)

Why the others are wrong

• (b)
  K₁ + K₂ = 0 — 2 + 6 = 8, not 0. K₁ + K₂ = 0 would need K₂ = −K₁, and the ratios make both constants positive.
• (c)
  K₂ = K₁ — K₂ = 6 but K₁ = 2, so they are not equal. The ratio 8⁄16 fixes each constant separately, and the two come out different.
• (d)
  K₂ + 3K₁ = 0 — 6 + 3 × 2 = 12, not 0. The multiplier 3 is right but the sign is wrong — this relation would need K₂ = −3K₁.

For a₁x + b₁y = c₁ and a₂x + b₂y = c₂:
a₁⁄a₂ = b₁⁄b₂ = c₁⁄c₂ → coincident lines, infinitely many solutions
a₁⁄a₂ = b₁⁄b₂ ≠ c₁⁄c₂ → parallel lines, no solution
a₁⁄a₂ ≠ b₁⁄b₂ → intersecting lines, a unique solution

When both constants are known, their ratio is the anchor: it fixes each unknown coefficient on its own.

Key facts

• Coincident lines: a₁⁄a₂ = b₁⁄b₂ = c₁⁄c₂.
• Here 8⁄16 = 1⁄2 gives K₁ = 2 and K₂ = 6.
• Coincident lines and infinitely many solutions describe the same system.

Common traps

• Stopping at K₁⁄4 = 3⁄K₂, which gives only K₁K₂ = 12 — a relation parallel lines satisfy too
• Inverting one ratio (4⁄K₁ beside 3⁄K₂), which scrambles the relation between K₁ and K₂

No directly related past PYQ was found.