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

Given:

Points $(3k – 1, k – 2), (k, k – 7)$ and $(k – 1, -k – 2)$ are collinear.

To do:

We have to find the value(s) of $k$.

Solution:

Let $A (3k-1, k-2), B (k, k-7)$ and $C (k-1, -k-2)$ be the vertices of $\triangle ABC$.

We know that,

Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by, 

Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Therefore,

Area of triangle \( ABC=\frac{1}{2}[(3k-1)(k-7+k+2)+k(-k-2-k+2)+(k-1)(k-2-k+7)] \)

\( 0=\frac{1}{2}[(3k-1)(2k-5)+k(-2k)+(k-1)(5)] \)

\( 0(2)=(6k^2-15k-2k+5-2k^2+5k-5) \)

\( 0=4k^2-12k \)

\( 4k(k-3)=0 \)

\( 4k=0 \) or \( k-3=0 \)

\( k=0 \) or \( k=3 \)

The values of $k$ are $0$ and $3$.