Question

Find the value of $k$ so that the line $3x-ky-1=0$ may be perpendicular to $4x-y-3=0$

Answer 1

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

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

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

Slope of line $\left(2\right)=-\frac{coefficientofx}{coefficientofy}=-\frac{4}{-1}=4$

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

$\Rightarrow \frac{3}{k}\times 4=-1$

$\Rightarrow \frac{12}{k}=-1$

$\Rightarrow 12=-k$

$\Rightarrow k=-12$