When considering system

\begin{equation*}

\mathbf{x}'=A\mathbf{x} \qquad \text{with } \

A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}

\end{equation*}

and discovering that it has a repeated root $\mu\ne 0$, but only one eigenvector, we know that $\mu>0$ corresponds to an unstable node, and $\mu<0$ t corresponds to a stable node. Furthermore, on the pictures below a straight line is directed along this single eigenvector. However, how to distinguish between two upper pictures and two lower pictures?

Observe that, $\mu$ is a double root of the equation $\lambda ^2- (a+d)\lambda + ad -bc=0$ with the discriminant $D:=(a+d)^2-4(ad -bc))=(a-d)^2+4bc$, and we consider the case $D=0\implies bc <0$ (except $a=d$). Thus $b$ and $c$ are not $0$ and have opposite signs.

Then, if $b<0$ (and $c>0$) "rotation" is counter-clockwise, and if $b>0$ (and $c<0$) "rotation" is clockwise. I say "rotation" because it is not a real rotation as in the

case of complex-conjugate roots, only a half-turn rotation by $\pm \pi$, but it works! Indeed, if we slightly increase $|b|$ or $|c|$, we get $D<0$ and we will have a focal point and the picture needs to be consistent.

If $a=d=\mu$ then either $b=0$ or $c=0$ but not both (otherwise there would be 2 linearly independent eigenvectors) and we use the same sign criteria, looking at $b$ or $c$ which is not $0$.