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The points (k + 1, 1), (2k + 1, 3) and (2k + 2, 2k) are collinear if
- a)k = - 1, 2
- b)
- c)k = 2, 1
- d)
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
The points (k + 1, 1), (2k + 1, 3) and (2k + 2, 2k) are collinear ifa)...
∵ Points are collinear.
∴ (k + 1) (3 - 2k) + (2k, + 1) (2k - 1) + (2k + 2) (1 - 3) = 0
⇒ 3k+3 - 2k2 - 2k + 4k2 - 1 - 4k - 4 = 0 ⇒ 2k2 - 3k - 2 = 0
⇒ 2k2 - 4k + k - 2 = 0
⇒ 2k(k - 2) + 1(k - 2) = 0
⇒ (2k + 1) (k - 2) = 0
∴ (k + 1) (3 - 2k) + (2k, + 1) (2k - 1) + (2k + 2) (1 - 3) = 0
⇒ 3k+3 - 2k2 - 2k + 4k2 - 1 - 4k - 4 = 0 ⇒ 2k2 - 3k - 2 = 0
⇒ 2k2 - 4k + k - 2 = 0
⇒ 2k(k - 2) + 1(k - 2) = 0
⇒ (2k + 1) (k - 2) = 0
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The points (k + 1, 1), (2k + 1, 3) and (2k + 2, 2k) are collinear ifa)...
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The points (k + 1, 1), (2k + 1, 3) and (2k + 2, 2k) are collinear ifa)...
Solve the problem by using the formula of area of triangle, taking the area as 0 (because it is a collinear line)
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The points (k + 1, 1), (2k + 1, 3) and (2k + 2, 2k) are collinear ifa)k = - 1, 2b)c)k = 2, 1d)Correct answer is option 'D'. Can you explain this answer?
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