= Solution
{c}
Write $u=(a,b)^T$, $v=(c,d)^T$ and $\Delta=ad-bc$. If $xu+yv=0$, then
$$
ax+cy=0,\qquad bx+dy=0.
$$
Eliminating $y$ gives $\Delta x=0$, and eliminating $x$ gives $\Delta y=0$. Hence $\Delta\ne0$ implies <linear independence>.
For the converse, suppose $\Delta=0$. If $u=0$, the family already contains the <zero vector> and is <linearly dependent>. If $a\ne0$, the equality $ad=bc$ gives $v=(c/a)u$. If $a=0$ but $u\ne0$, then $b\ne0$ and $bc=0$ forces $c=0$, so $v=(d/b)u$. Both cases give a nontrivial <linear relation>. Therefore
$$
\boxed{u,v\text{ are linearly independent}\iff ad-bc\ne0.}
$$
This is the two-dimensional <determinant> criterion, proved directly without assuming that criterion.
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