Question

Find the value of k so that the line $kx-2ky-3=0$ may be perpendicular to $2x-3ky-4=0$

Answer 1

Given:$kx-2ky-3=0$ .......$\left(1\right)$

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

Given:$2x-3ky-4=0$ .......$\left(2\right)$

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

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

$\Rightarrow \frac{1}{2}\times \frac{2}{3k}=-1$

$\Rightarrow 1=-3k$

$\therefore k=\frac{-1}{3}$