Question

Find the value of $k$ so that the line $x+2ky+1=0$ may be perpendicular to $x-2ky-2=0$

Answer 1

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

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

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

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

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

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

$\Rightarrow 4k^2=1$

$\Rightarrow k^2=\frac{1}{4}$

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