Question

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

• A. 1
• B. 0
• C. -1
• D. 2

Answer 1

The correct option is B 0

Given,

$2x+3y+4=0$ and $\lambda$$x+ky+2=0$

multiplying 2nd equation and comparing with 1st,

$2\lambda =2$ and $2k=3$

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

Now,

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

$=>0$