Question

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

Answer 1

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

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

Given:$x-3y-1=0$ .......$\left(2\right)$

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

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

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

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

$\Rightarrow 1=-9k$

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