Proof involving homogeneous system of linear equations with det 0.

This is from Hoffman & Kunze:

Consider $Ax = 0$ with $A = \bigl(\begin{smallmatrix} a&b\\ c&d \end{smallmatrix} \bigr)$. Such that $ad - bc = 0$ With some element of A nonzero.

Then there is a solution $(x^0_1, x^0_2)$ such that $(x_1,x_2)$ is a solution if and only if there is a scalar $y$ such that $x_1 = yx_1^0$ and $x_2 = yx_2^0$.

Some stuff I've noticed

1. $ad-bc=0$ is the determinant of the matrix.
2. if $ad = 0$ then $bc = 0$ so there is at least the cases $ad=0$ & $bc = 0$ or $ad\neq0$ and $bc = ad$

Unfortunately however I am not seeing how to setup the problem for a proof.

Can anybody out there provide a hint as to how to proceed?

EDIT: Is it enough to say without loss of generality assume $a \neq 0$ and then $\bigl(\begin{smallmatrix} a&b\\ c&d \end{smallmatrix} \bigr)$ can be transformed to $\bigl(\begin{smallmatrix} 1&\frac{b}{a}\\ c&d \end{smallmatrix} \bigr)$ can be transformed to $\bigl(\begin{smallmatrix} 1&\frac{b}{a}\\ c-c&d-\frac{bc}{a} \end{smallmatrix} \bigr)$ which equals $\bigl(\begin{smallmatrix} 1&\frac{b}{a}\\ 0&0 \end{smallmatrix} \bigr)$ by the assumptions made. so $x_1 = -\frac{b}{a}x_2$ and multiplication by $y$ does not change the fact that it is a solution? Is this enough to solve the whole problem or is there other cases I need to check?
If one of the entries is non-zero, then $A$ cannot be the zero transformation. Therefore the dimension of the range is at least $1$. If the determinant is $0$, then the dimension of the range is less than $2$. Therefore it is equal to $1$. Together this says that there are two linearly independent vectors, $v_1$, $v_2$, such that $Av_1$ is non-zero and $Av_2$ is the zero vector. Since these two vectors are linearly independent, any vector in the domain is a linear combination of them. Thus $Av=0$ if an only if $v=kv_2$ for some $k$.