Question

3.
If (k, 2 - 2k), (-k + 1, 2k), (-4 - k, 6 - 2k) are collinear, then k =
1)2
2)5
3)-1
4)3​

Answer 1

Answer:

k= -1

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Answer 2

Answer:

Given points (k, 2 – 2k), (-k + 1, 2k) and (-4 – k, 6 – 2k) are collinear. ⇒ k [2k – ( 6 – 2k)] – (2 – 2k) [(-k + 1) – (-4 – k)] + 1 [(-k + 1) (6 – 2k) – (-4 – k) (2k)] = 0 ⇒ k [2k – 6 + 2k] – (2 – 2k) [-k + 1 + 4 + k ] + 1[- 6k + 2k2 + 6 – 2k + 8k + 2k2] = 0 ⇒ K(4k – 6) – (2 – 2k) (5) + 1(4k2 + 6) = 0 ⇒ 4k2 – 6k – 10 + 10k + 4k2 + 6 = 0 ⇒ 8k2 + 4k – 4 = 0 ⇒ 2k2 + k – 1 = 0 ⇒ 2k 2 + (2 – 1)k – 1 = 0 ⇒ 2k(k + 1) – 1(k + 1) = 0 ⇒ (k + 1) (2k -1) = 0 Hence, k = -1, 1/2.

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