Question

The lines $x+y=1,$ $(k-1)x+(k^2-7)y-5=0$ and $(k-2)x+(2k-5)y=0$ are

• A. concurrent for three distinct values of $k$
• B. concurrent for exactly one value of $k$
• C. concurrent for no value of $k$
• D. For $k=3,$ the lines are concurrent

Answer 1

The correct option is C concurrent for no value of $k$

For concurrency of three lines, $\Delta =0$

$\Rightarrow \left|\begin{array}{ccc}1& 1& -1\\ k-1& k^2-7& -5\\ k-2& 2k-5& 0\end{array}\right|=0$

$\Rightarrow k^3-4k^2+5k-6=0$
$\Rightarrow (k-3)(k^2-k+2)=0$
$\Rightarrow k=3$ and $k^2-k+2=0$ has no real roots.
But for $k=3,$ the lines are parallel.