For what value of k,

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), $\frac{3}{6}=\frac{-1}{-k}=\frac{8}{16}$

$\Rightarrow$ $\frac{1}{k}=\frac{1}{2}$

$\therefore$ $k=2$

 

 

 

 

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