Question

Find the value of $k$ so that the line $4x+ky-11=0$ may be perpendicular to $4x+5y-3=0$

Answer 1

Given:$4x+ky-11=0$ .......$\left(1\right)$

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

Given:$4x+5y-3=0$ .......$\left(2\right)$

Slope of line $\left(2\right)=-\frac{coefficientofx}{coefficientofy}=-\frac{4}{5}$

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

$\Rightarrow \frac{-4}{k}\times \frac{-4}{5}=-1$

$\Rightarrow \frac{16}{5k}=-1$

$\Rightarrow 16=-5k$

$\Rightarrow k=\frac{-16}{5}$