The points (k, –2k), (–k + 1, 2k) and
(–4 –k, 6 –2k) can't be collinear for any value of k.

True

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False

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Solution

The correct option is A True
If the three points are collinear, then area of triangle formed by these points will be zero.
$i . e ., \frac{1}{2} ⎛ ⎜ ⎝ \begin{matrix} 1 & k & - 2 k 1 & - k + 1 & 2 k 1 & - 4 - k & 6 - 2 k \end{matrix} ∣ ∣ ∣ ∣ = 0 . . . . . . . (1] i . e ., \frac{1}{2} (A | = 0, say On applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1} on A, we get (1] as ⎛ ⎜ ⎝ \begin{matrix} 1 & k & - 2 k 0 & - 2 k + 1 & 4 k 0 & - 4 - 2 k & 6 \end{matrix} ∣ ∣ ∣ ∣ = 0 Expanding along R_{1}, we get - 12 k + 6 + 4 k (4 + 2 k) = 0 or 8 k^{2} + 4 k + 6 = 0 or 4 k^{2} + 2 k + 3 = 0 For given equation, D = {(2)}^{2} - 4 \times 4 \times 3 < 0 Hence, real roots of this equation does not exist . So, given points can' t be collinear for any value of k .$

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

k =

k

(k, 2 - 2 k)

(1 - k, 2 k)

(- 4 - k, 6 - 2 k)

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