Question

$2x+3y+4=0$ & $\lambda x+ky+2=0$ identical lines then

• A. $\lambda =1,k=\frac{1}{2}$
• B. $\lambda =1,k=\frac{3}{2}$
• C. $\lambda =1,k=\frac{5}{2}$
• D. $\lambda =-1,k=\frac{1}{2}$

Answer 1

The correct option is B $\lambda =1,k=\frac{3}{2}$

Given

$2x+3y+4=0\cdots \left(1\right)$

$\lambda x+ky+2=0\cdots \left(2\right)$

According to Question

$\left(1\right),\left(2\right)$ are identical

$\left(1\right)....\times \frac{1}{2}$

$⟹x+\frac{3}{2}y+2=0$

$⟹x+\frac{3}{2}y+2=\lambda x+ky+2$

Comparing coeffiecients

$\lambda =1k=\frac{3}{2}$