Question

Find the value of $k$ so that the line $x+3ky+1=0$ may be perpendicular to $x-3ky-6=0$

Answer 1

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

Slope of line $\left(1\right)=-\frac{coefficientofx}{coefficientofy}=-\frac{1}{3k}$

Given:$x-3ky-6=0$ .......$\left(2\right)$

Slope of line $\left(2\right)=-\frac{coefficientofx}{coefficientofy}=-\frac{1}{-3k}=\frac{1}{3k}$

As the lines $\left(1\right)$ and $\left(2\right)$ are perpendicular, the product of their slopes$=-1$

$\Rightarrow -\frac{1}{3k}\times \frac{1}{3k}=-1$

$\Rightarrow \frac{-1}{9k}=-1$

$\Rightarrow 1=9k$

$\Rightarrow k=\frac{1}{9}$