If the points $(k, 2 k), (3 k, 3 k)$ and $(3, 1)$ are collinear then the value of $k$ is

\frac{7}{9}

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\frac{2}{3}

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\frac{- 2}{3}

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\frac{- 1}{3}

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Solution

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero
The area of the triangle with vertices $(x_{1}, y_{1})$ ; $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is $∣ ∣ ∣ \frac{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})}{2} ∣ ∣ ∣$

Hence, substituting the points
$(x_{1}, y_{1}) = (k, 2 k)$ ; $(x_{2}, y_{2}) = (3 k, 3 k)$ and $(x_{3}, y_{3}) = (3, 1)$ in the area formula, we get
$∣ ∣ ∣ \frac{k (3 k - 1) + 3 k (1 - 2 k) + 3 (2 k - 3 k)}{2} ∣ ∣ ∣ = 0$

$\Rightarrow 3 k^{2} - k + 3 k - 6 k^{2} + 6 k - 9 k = 0$
$\Rightarrow - 3 k^{2} - k = 0$
$\Rightarrow - k (3 k + 1) = 0$
$\Rightarrow k = 0, 3 k + 1 = 0$
$k = 0$ or $k = - \frac{1}{3}$
Hence, Option D is correct.

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