Question

If the line $(\frac{x}{2}+\frac{y}{3}-1)+\lambda (2x+y-1)=0$ is parallel to x-axis then $\lambda =$

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $-\frac{1}{4}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $-\frac{1}{4}$

$(\frac{x}{2}+\frac{y}{3}-1)+\lambda (2x+y-1)=0$

$(3x+2y-6)+6\lambda (2x+y-1)=0$

$(3+12\lambda )x+(2+6\lambda )y-(6+6\lambda )=0$

$m=0$

$-\left(\frac{3+12\lambda }{2+6\lambda }\right)=0$

$3+12\lambda =0$

$\lambda =\frac{-3}{12}=\frac{-1}{4}$