TITLE: For what values of $a$ will the lines $ay+3x=4$ and $2y+4x=3$ have no intersection points?
QUESTION [5 upvotes]: Multiple choice question:

For what values of $a$ will the lines $ay+3x=4$ and $2y+4x=3$ have no intersection points?

A)  2
B) 1.5
C) 8/3
D) -0.5
E) -2
  
  


I tried rearranging them into $y=mx+b$ form and solve, but I will end up with a literal answer like this
$$a=\frac{(4-3x)}{y}$$
How do you work this out? I think you need to make the gradient the same but how do you do that?

REPLY [2 votes]: The system is equivalent to
$$
\begin{bmatrix}
3&a\\
4&2
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=\begin{bmatrix}
4\\
3
\end{bmatrix}.
$$
Let $A$ be the $2\times 2$ matrix above. The lines don't intersect iff $(4,
3)^T$ is not in the span of the columns of $A$. The lines don't intersect iff the columns are linearly dependent (as $(4,3)^T\notin\text{span} \{(3,4)^T$}. Thus the lines don't intersect iff
$$
\lambda
\begin{bmatrix}
a\\
2
\end{bmatrix}
=
\begin{bmatrix}
3\\
4
\end{bmatrix}
$$
for some $\lambda\in\mathbb{R}$ whence $\lambda a=3$ i.e. $a=3/2$.