Question

The number of real values of $\lambda$ for which the lines $x-2y+3=0,\lambda x+3y+1=0$ and $4x-\lambda y+2=0$ are concurrent is

• A. 0
• B. 1
• C. 2
• D. Infinite

Answer 1

The correct option is A

0

The given lines are

$x-2y+3=0...\left(1\right)$

$\lambda x+3y+1=0...\left(2\right)$

$4x-\lambda y+2=0...\left(3\right)$

It is given that (1), (2) and (3) are concurrent.

$\therefore \left|\begin{array}{ccc}1& -2& 3\\ \lambda & 3& 1\\ 4& -\lambda & 2\end{array}\right|=0$

$\Rightarrow (6+\lambda )+2(2\lambda -4)+3(-{\lambda }^2-12)=0$

$\Rightarrow 6+\lambda +4\lambda -8-3{\lambda }^2-36=0$

$\Rightarrow 5\lambda -3{\lambda }^2-38=0$

$\Rightarrow 3{\lambda }^2-5\lambda +38=0$

The discriminant of this equation is

$25-4\times 3\times 38=-431$

Hence, there is no real value of 1 for which the lines $x-2y+3=0,\lambda x+3y+1=0$ and $4x-\lambda y+2=0$ are concurrent.