Next level of resistor divider
Two main factors are how much current can you draw (that is the voltage applied divided by the total resistance - sum of the two resistors. The second factor is what output impedance you want (or can tolerate) from the divider point. That value is the parallel combination of the two resistors if it's being fed from a stiff voltage source. 

So in your example, assume 10V is applied, the 100K + 100K divider draws 50uA and has a source impedance of 100K||100K = 50K ohms (so if you load it with 50M ohms it will drop by about 0.1%)

In the case of the 10R + 10R, it will draw 500mA and have a source impedance of 5 ohms, so if you load it with 5K ohms it will drop about 0.1%. 

The other practical factor if you still have some range in the above two constraints is that it's nice to be able to use commercially available resistor values. Maybe you can change the total resistance a bit to get it to fit with values that you can buy in the tolerance and power rating you need. If you can't buy the exact values you calculate and can't seem to make it fit, another approach is to pick one of the two values that you can buy and then find a parallel or series combination of standard values that makes the ratio correct (since the ratio is usually much more important than the total resistance). 

For example, suppose you want to divide 12V down to 2.5V so you can measure it with a microcontroller, and the microcontroller can only tolerate 10K source impedance before errors increase. 

So you might pick 38K and 10K, which has a source impedance of 7.92K, but 38K isn't a standard E96 value- the closest is 38.3K, which is an error of about 0.6%. You could also use 5.62K and 21.3K which would have a ratio error of only about 0.2% but would draw more current from the 12V input (and would still meet the 10K max source impedance limit, of course). 

Edit: Given your added comment about AC signals, if you match the resistance (actually you match conductance- reciprocal of resistance) with capacitance across the divider you can maintain the same shape to the signals. Since resistors are not designed to have any specific capacitance, only stray and trace capacitance, you might have to adjust the capacitance across one of the resistors. That's essentially what you do when you adjust your 10:1 oscilloscope probe compensation.