Conditions for two straight lines to intersect: is this exam question wrong? I am pretty sure this question (from a university admission test exam) is wrong.

Two lines: $a_1x+b_1y+c_1=0$, $a_2x+b_2y+c_2=0$, intersect only if
(a) $a_1a_2-b_1b_2=0\;\;\;$ (b) $a_1a_2-b_1b_2\ne0\;\;\;$(c) $a_1a_2-b_1b_2=1\;\;\;$ (d) $a_1a_2-b_1b_2\ne1$

Attempted solution:
\begin{matrix}
a_1x+b_1y+c_1=0&                    \;\; &a_2x+b_2y+c_2=0 \\
y=-\frac{a_1x}{b_1}-\frac{c_1}{b_1}&\;\; & y=-\frac{a_2x}{b_2}-\frac{c_2}{b_2}\\
\end{matrix}
If the lines are not parallel, they will always intersect. In case of non-parallel lines 
$$m_1\ne m_2$$
$$-\frac{a_1}{b_1}\ne-\frac{a_2}{b_2}$$
$$\require{cancel}
\cancel{-}a_1b_2\ne\cancel{-}a_2b_1$$
$$a_1b_2-a_2b_1\ne0.$$
 A: If two lines do not intersect--i.e., they have no points in common--then the system of equations $$\begin{align*} a_1 x + b_1 y + c_1 &= 0 \\ a_2 x + b_2 y + c_2 &= 0 \end{align*}$$ will have no solution for $(x,y)$.  Thus, if we solve the system, we find $$x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}, \quad y = \frac{a_2 c_1 - a_1 c_2}{a_1 b_2 - a_2 b_1}.$$  This solution does not exist or is indeterminate if $a_1 b_2 - a_2 b_1 = 0$.
However, some care is required:  two lines coincide if $$(a_1, b_1, c_1) = k(a_2, b_2, c_2)$$ for some nonzero scalar constant $k$, and in this case, both the numerators and denominator of the aforementioned solution are zero, meaning that there are infinitely many points that the two equations share in common.  So while it is a sufficient condition for $a_1 b_2 - a_2 b_1 \ne 0$ to imply that the lines intersect, it is not a strictly necessary condition, and for that reason, the question should have been better phrased by saying "two lines...intersect if..." rather than "only if"; alternatively, it might be phrased "two distinct lines...intersect only if...."
