Question

If the points (k,2-k),(-k+1,2k),(-4-k,6-2k) are collinear then find k.

Answer 1

Answer:

Given points are collinear.So,Area=0

1/2[k(2k-6+2k)+(-k)(6-2k-2+k)+(-4-k)(2-k-2k)]=0

[k(4k-6)+(-k)(4-k)+(-4-k)(2-3k)]=0

[4k^2 - 6k -4k+ k^2 -8+ 12k- 2k+ 3k^2]=0

8k^2-8=0

8k^2=8

k^2=1

k=1 or k=(-1)

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