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Question: If the points \((k,2 - 2k)\), \((1 - k,\ 2k)\) and \(( - k - 4,\ 6 - 2k)\) be collinear, then the po...

If the points (k,2−2k)(k,2 - 2k), (1−k, 2k)(1 - k,\ 2k) and (−k−4, 6−2k)( - k - 4,\ 6 - 2k) be collinear, then the possible values of k are.

A

12,−1\frac{1}{2}, - 1

B

1,−121, - \frac{1}{2}

C

1,−21, - 2

D

2,−12, - 1

Answer

12,−1\frac{1}{2}, - 1

Explanation

Solution

The points are collinear if the area of triangle formed by these three points is zero.

⇒12[k{2k−(6−2k)}+(1−k){(6−2k)−(2−2k)}\Rightarrow \frac { 1 } { 2 } [ k \{ 2 k - ( 6 - 2 k ) \} + ( 1 - k ) \{ ( 6 - 2 k ) - ( 2 - 2 k ) \} +(−4−k){(2−2k)−2k}]=0+ ( - 4 - k ) \{ ( 2 - 2 k ) - 2 k \} ] = 0

On simplification, we get k=−1k = - 1 or 12\frac { 1 } { 2 }.