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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
For lines to be concurrent they must pass through a common point.
x – y – 1 = 0
2x + 3y – 12 = 0
2x – 3y + k = 0
x – y = 1 × 2
2x + 3y = 12
2x – 2y = 12
5y = 10
y = 2
x – y = 1
x – 2 = 1
x = 3
Then 2x – 3y + k = 0 also passes from (3, 2)
2 (3) – 3 (2) + K = 0
6 – 6 + K = 0
K = 0