Question

The product of all the values of $\left|\lambda \right|$ , such that the lines $x+2y-3=0,3x-y-1=0$ and $\lambda x+y-2=0$ cannot form a triangle, is equal to $k$ . Find $10k$

Answer 1

Answer: 18

Case I: $x+2y-3=0$ and $\lambda x+y-2=0$ are parallel
$\Rightarrow \frac{\lambda }{1}=\frac{1}{2}\Rightarrow \left|\lambda \right|=0.5$
Case II: $3x-y-1=0$ and $\lambda x+y-2=0$ are parallel
$\Rightarrow \frac{\lambda }{3}=\frac{1}{-1}\Rightarrow \left|\lambda \right|=3$
Case III: Lines $x+2y-3=0,3x-y-1=0$ and $2\lambda x+y-2=0$ are concurrent
$\Rightarrow \left|\begin{array}{ccc}\lambda & 1& -2\\ 1& 2& -3\\ 3& -1& -1\end{array}\right|=0\Rightarrow -5\lambda -8+14=0\Rightarrow \left|\lambda \right|=1.2$
Hence, the product of all the values of $\left|\lambda \right|$ $=0.5\times 3\times 1.2=1.8$