MCQ
If $2x + 3y + 4 = 0 \ \& \ \text{amp}; \lambda\text{x}+\text{ky}+2=0$ are identical lines then $3\lambda-2\text{k}=$
- A$1$
- ✓$0$
- C$-1$
- D$2$
✓
Answer
Correct option: B.
$0$
Given,$ 2x + 3y + 4 = 0$ and $\lambda\text{x}+\text{ky}+2=0$
multiplying $2^{nd}$ equation and comparing with $1^{st},$
$2\lambda=2$ and $2\text{k}=3=>\lambda=1$ and $\text{k}=\frac{3}{2}$
Now $,3\lambda-2\text{k}=3\times1-2\times\frac{3}{2}= > 0$
multiplying $2^{nd}$ equation and comparing with $1^{st},$
$2\lambda=2$ and $2\text{k}=3=>\lambda=1$ and $\text{k}=\frac{3}{2}$
Now $,3\lambda-2\text{k}=3\times1-2\times\frac{3}{2}= > 0$
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