Question:
For what value of $k$, do the equations $3 x-y+8=0$ and $6 x-k y=-16$
(a) $\frac{1}{2}$
(b) $-\frac{1}{2}$
(c) 2
(d) $-2$
Solution:
(c) Condition for coincident lines is
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Given lines. $3 x-y+8=0$
and $6 x-k y+16=0$
Here, $a_{1}=3, b_{1}=-1, c_{1}=8$
and $a_{2}=6, b_{2}=-k, c_{2}=16$
From Eq. (i), $\frac{3}{6}=\frac{-1}{-k}=\frac{8}{16}$
$\Rightarrow$ $\frac{1}{k}=\frac{1}{2}$
$\therefore$ $k=2$