Question

If the straight line $3x+4y+5-k(x+y+3)=0$ is parallel to $y-$ axis, then the value of $k$ is

Answer 1

$3x+4y+5-k(x+y+3)=0$
$\Rightarrow (3-k)x+(4-k)y+5-3k=0$
$\Rightarrow y=\left(\frac{k-3}{4-k}\right)x+\frac{3k-5}{4-k}$
Here it is parallel to $y-$ axis
Therefore $\tan \theta =m=\frac{k-3}{4-k}$ is not defined
So $k-4=0\Rightarrow k=4$

Alternatively
A straight line is parallel to $y-$axis is of form $x=k$
$\therefore$ its $y$ coefficient will be zero.
$3x+4y+5-k(x+y+3)=0$
$\Rightarrow (3-k)x+(4-k)y+5-3k=0$
$\Rightarrow 4-k=0\Rightarrow k=4$