For what value of K does the following system have an infinite number of solutions? x + 2Ky - 8 = 0 2x - y - 16 = 0

Correct — C. For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to have infinitely many solutions, all three coefficient ratios must agree.

a₁⁄a₂ = 1⁄2
c₁⁄c₂ = (−8)⁄(−16) = 1⁄2 — those two already match
b₁⁄b₂ = 2K⁄(−1) must equal 1⁄2 as well

2K × 2 = −1 → 4K = −1
K = −1⁄4, the value printed on option (c).

Why the others are wrong

• (a)
  Option (a) is 4. Put K = 4 and the y-coefficients become 8 and −1, a ratio of −8 against a₁⁄a₂ = 1⁄2. The lines then cross at a single point.
• (b)
  Option (b) is −1⁄2, making 2K = −1 and the y-ratio (−1)⁄(−1) = 1. That is not 1⁄2, so the two equations are not the same line.
• (d)
  Option (d) is 2, making 2K = 4 and the y-ratio 4⁄(−1) = −4. The three ratios disagree again, so the system has one solution rather than infinitely many.

Two straight lines can meet once, never, or everywhere. Written as a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, three ratios decide which:

a₁⁄a₂ ≠ b₁⁄b₂ — they cross once, one solution.
a₁⁄a₂ = b₁⁄b₂ ≠ c₁⁄c₂ — parallel, no solution.
a₁⁄a₂ = b₁⁄b₂ = c₁⁄c₂ — the same line twice, infinitely many solutions.

Both equations here are already in that form, with c₁ = −8 and c₂ = −16, so the constants are negative and their ratio is the positive 1⁄2.

Key facts

• Infinitely many solutions require a₁/a₂ = b₁/b₂ = c₁/c₂.
• Equal a and b ratios with a different c ratio means parallel lines and no solution at all.
• Both equations must be arranged as ax + by + c = 0 before the coefficients are read off.
• Here a₁/a₂ = 1/2 and c₁/c₂ = (−8)/(−16) = 1/2, which forces b₁/b₂ = 2K/(−1) to be 1/2.

Common traps

• Reading the constants as +8 and +16 when the paper prints −8 and −16.
• Checking only a₁/a₂ = b₁/b₂, which is equally true of parallel lines with no solution.
• Solving 2K/(−1) = 1/2 as 4K = 1 and losing the minus sign.

No directly related past PYQ was found.