Question

Find the value of $k$ so that the line $x+5ky-3=0$ may be parallel to $x+y-6=0$

Answer 1

Given:$x+5ky-3=0$ .......$\left(1\right)$

Slope of line $\left(1\right)=-\frac{coefficientofx}{coefficientofy}=-\frac{1}{5k}$

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,$-\frac{1}{5k}=-1$

$\Rightarrow 5k=1$

$\therefore k=\frac{1}{5}$