The value of k for which the points 2-3- k -1 and 0-4 are collinear- is -A- k -107B- k -107C- k -187D- k -187

The correct option is B k=107If the given points are collinear then, 2 31k 11041=0On nbsp;applying nbsp;R2→R2 R1and nbsp;R3→R3 R1, nbsp;we nbsp;get 2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Area of Triangle with Coordinates of Vertices Given

The value of ...

Question

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

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

The correct option is B k=107If the given points are collinear then, ⎛⎜⎝2−31k−11041∣∣ ∣∣=0On applying R2→R2−R1and R3→R3−R1, we get−⎛⎜⎝2−31k−220−270∣∣ ∣∣=0On expanding along C3 we get−(k−22−27∣∣∣=07k−14+4=0⇒k=107