- A0
- B3
- C−3
- D6
Step-by-Step Solution
Collinear points enclose zero area: ½|2(k + 3) + 4(−3 − 3) + 6(3 − k)| = 0. This gives |2k + 6 − 24 + 18 − 6k| = |−4k| = 0, so k = 0.
Answer: A) 0
The three points (2, 3), (4, k) and (6, −3) lie on one straight line. Find the value of k.
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