find the value of k for which the points
( 3k-1 , k-2 ) , ( k , k- 7 ) and ( k-1 , -k-2) are collinear
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If the given points are collinear then the area of triangle formed by the points should be equal to zero.
Area of triangle formed by the points,
⇒1/2 {(3k-1)(2k+5) + (k)(-2k) + (k-1)(5)} = 0
On further simplification we get,
⇒ ( 6k² - 17k + 5 -2k² +5k - 5) = 0
⇒ 4k² - 12k = 0
⇒ 4k (k - 3) = 0
⇒ k - 3 = 0
⇒ k = 3
Hence the value of k is 3.
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