The value of k for which the points (2,-3), (k,-1) and (0,4) are collinear, is .

k = - \frac{10}{7}

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k = \frac{10}{7}

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k = \frac{18}{7}

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k = - \frac{18}{7}

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Solution

The correct option is B $k = \frac{10}{7}$
If the given points are collinear then,
$⎛ ⎜ ⎝ \begin{matrix} 2 & - 3 & 1 k & - 1 & 1 0 & 4 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0 On applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}, we get - ⎛ ⎜ ⎝ \begin{matrix} 2 & - 3 & 1 k - 2 & 2 & 0 - 2 & 7 & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0 On expanding along C_{3} we get - (\begin{matrix} k - 2 & 2 - 2 & 7 \end{matrix} ∣ ∣ ∣ = 0 7 k - 14 + 4 = 0 \Rightarrow k = \frac{10}{7}$

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