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

k = - \frac{10}{7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

k = \frac{10}{7}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

k = \frac{18}{7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

k = - \frac{18}{7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

k

(2, 3), (3, k)

(3, 7)

Find the value(s) of $k$ for which the points $(3 k - 1, k - 2), (k, k - 7)$ and $(k - 1, - k - 2)$ are collinear.

Join BYJU'S Learning Program