Question

The line $\left(3x-y+5\right)+\lambda \left(2x-3y-4\right)=0$ will be parallel to $y$-axis, if $\lambda =$

• A. $\frac{1}{3}$
• B. $-\frac{3}{2}$
• C. $-\frac{1}{3}$
• D. $\frac{3}{2}$

Answer 1

The correct option is C

$-\frac{1}{3}$

Calculate the value of $\lambda$.

The given equation is $\left(3x-y+5\right)+\lambda \left(2x-3y-4\right)=0$

$\begin{array}{rcl}& \Rightarrow & 3x-y+5+2\lambda x-3\lambda y-4\lambda =0\\ & \Rightarrow & x\left(3+2\lambda \right)-y\left(1+3\lambda \right)+5-4\lambda =0\end{array}$

Here slope of the line is $m=\frac{3+2\lambda }{1+3\lambda }$

Since the line is parallel to $y$ axis, slope is undefined.

$\begin{array}{rcl}& \Rightarrow & 1+3\lambda =0\\ & \Rightarrow & \lambda =-\frac{1}{3}\end{array}$

Hence option $\left(C\right)$ is correct.