Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $infinite$

Answer 1

The correct option is A $0$

For concurrency of lines $a_{1}x+b_{1}y+c_1=0$, $a_{2}x+b_{2}y+c_2=0$ and $a_{3}x+b_{3}y+c_3=0$, the condition is

$\left[\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right]=0$

So, For given lines condition for concurrency is,

$\left[\begin{array}{ccc}1& -2& 3\\ k& 3& 1\\ 4& -k& 2\end{array}\right]=0$

Evaluating we get,

$\to (6+k)+(4k-3k^2)+4(-2-9)=0$

$\to 3k^2-5k+38=0$

But for this equation $D=25-4\times 3\times 38<0$. So the given equation has no real root.

Thus, A is the correct answer.