Question

Find the value of $k$ for which the points $(4k-1,k-2),(k,k-7)$ and $(k-1,-k-2)$ are collinear.

Answer 1

For the given points $(4k-1,k-2),(k,k-7)$ and $(k-1,-k-2)$ to be collinear we must have the area of triangle formed by the points $(4k-1,k-2),(k,k-7)$ and $(k-1,-k-2)$ be zero.

This will give,

$\frac{1}{2}\left|\begin{array}{ccc}1& 4k-1& k-2\\ 1& k& k-7\\ 1& k-1& -k-2\end{array}\right|=0$

or, $\left|\begin{array}{ccc}1& 4k-1& k-2\\ 1& k& k-7\\ 1& k-1& -k-2\end{array}\right|=0$

[$R_2^{\prime }=R_2-R_1$ and $R_3^{\prime }=R_3=R_2$]

or, $\left|\begin{array}{ccc}1& 4k-1& k-2\\ 0& 1-3k& -5\\ 0& -1& 5-2k\end{array}\right|=0$

or, $(3k-1)(2k-5)-5=0$

or, $6k^2-17k=0$

or, $k(6k-17)=0$

or, $k=0,\frac{17}{6}$.

For the above values of $k$ the given three points will be co-linear.