Mathews

answers: 1

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Answer:

Consider the given points.

(k,−1),(2,1) and (4,5)

Since, these points are collinear means that the area of triangle must me zero.

So,

2

1

∣x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)∣=0

where (x

1

,y

1

),(x

2

,y

2

),(x

3

,y

3

) are the points

Therefore,

k(1−5)+2(5+1)+4(−1−1)=0

k(−4)+2(6)+4(−2)=0

−4k+12−8=0

−4k+4=0

4k=4

k=1

Hence, this is the answer.

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