Question

Find the value of $k$ so that the line $5x+ky-3=0$ may be parallel to $3x-2y-6=0$

Answer 1

Given:$5x+ky-3=0$ .......$\left(1\right)$

Slope of line $\left(1\right)=-\frac{coefficientofx}{coefficientofy}=-\frac{5}{k}$

Given:$3x-2y-6=0$ .......$\left(2\right)$

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

As the lines $\left(1\right)$ and $\left(2\right)$ are parallel,$-\frac{5}{k}=\frac{3}{2}$

$\Rightarrow 3k=-10$

$\therefore k=\frac{-10}{3}$