Question

Find the value of $k$ for which the lines $x+2y+3=0$ and $8x+ky-1=0$ are parallel.

Answer 1

We have

x+2yy+3=02x−by+5=0

and, 8x+ky-1=ax+3y=2

Now,

x+2y+3=02x−by+5=0

or, 2y=-x-3by=2x+5

or $y=\frac{-x}{2}-\frac{3}{2}$

Slope of the line $=-\frac{2}{b}$

Again

$8x+ky-1=0$

or $ky=-8x+1$

or $y=\frac{-8}{R}+\frac{1}{R}$

Solpe of the is line $=\frac{-8}{R}$

$\because$ The lines are parallel, so the slopes of the two lines are equal

$\therefore \frac{-1}{2}=\frac{-8}{R}orR=16$

Hence, which is the required answer.