Question

Find the value of $k$ so that the line $x+2ky-6=0$ may be parallel to $x-y-6=0$

Answer 1

Given:$x+2ky-6=0$ .......$\left(1\right)$

Slope of line $\left(1\right)=-\frac{coefficientofx}{coefficientofy}=-\frac{1}{2k}$

Given:$x-y-6=0$ .......$\left(2\right)$

Slope of line $\left(2\right)=-\frac{coefficientofx}{coefficientofy}=-\frac{1}{-1}=1$

As the lines $\left(1\right)$ and $\left(2\right)$ are parallel,$-\frac{1}{2k}=1$

$\Rightarrow 2k=-1$

$\therefore k=\frac{-1}{2}$