Question

Find the value of $\lambda$ such that the straight line $(2x+3y+4)+\lambda (6x-y+12)=0$ is parallel to $y-$axis.

Answer 1

Slope of the given line $=\frac{2+6\lambda }{3-\lambda }=\infty \Rightarrow \lambda =3$
We may also say that since the lines is parallel to $y-$axis, the coefficient of $y$ in its equation $P+\lambda Q=0$ must be zero. Therefore $3-\lambda =0$ or $\lambda =3$.