The points $(k, 2 - 2 k)$ , $(- k + 1, 2 k)$ and $(- 4 - k, 6 - 2 k)$ are collinear for

all values of k

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k = - 1

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k = 1 / 2

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no value of k.

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Solution

The correct options are
B $k = - 1$
C $k = 1 / 2$
The given points are collinear if

$∣ ∣ ∣ ∣ \begin{matrix} k & 2 - 2 k & 1 - k + 1 & 2 k & 1 - 4 - k & 6 - 2 k & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} k & 2 - 2 k & 1 - 2 k + 1 & 4 k - 2 & 0 - 4 - 2 k & 4 & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0$ $[R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}]$

$\Rightarrow 4 (- 2 k + 1) - (- 4 - 2 k) (4 k - 2) = 0$

$\Rightarrow (1 - 2 k) (4 - 8 - 4 k) = 0$

$\Rightarrow (1 - 2 k) (k + 1) = 0$

$\Rightarrow k = - 1 o r k = 1 / 2$

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Similar questions

For what value of k are the points $(k, 2 - 2 k), (- k + 1, 2 k), (- 4 - k, 6 - 2 k)$ collinear?

k

(k, 2 - 2 k)

(1 - k, 2 k)

(- 4 - k, 6 - 2 k)

(k, 2 - 2 k), (- k + 1, 2 k), (- 4 - k, 6 - 2 k)

k =

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