The value of k for which the following lines
x − y − 1 = 0 2x + 3y − 12 = 0 2x − 3y + k = 0
are concurrent is
Explanation:
Three lines are cocurrent when they meet at a single point i.e., the equations representing these lines have a common solution.
Consider, x − y − 1 = 0 ...(1) and 2x + 3y − 12 = 0 ...(2)
(2) + 3 × (1) ⇒ 5x -15 = 0 ⇒ x = 3
Putting x = 3 in (1) we get y = 2 ∴ (x, y) = (3, 2) is the common solution for first two lines.
Hence, (3, 2) should also satisfy the 3rd equation.
⇒ 2 × 3 − 3 × 2 + k = 0 ⇒ k = 0
Hence, option (c).
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