light led at specific resistance?

90 ohms by itself won't be enough to change the voltage, but 90 ohms paried with a second resistor would.  Most systems that use a resistive based sensor use what is called a "voltage divider".  The voltage divider is two resistors wired in series from 12vdc to ground.  The output voltage between the two resistors can vary drastically as the resistance of one resistor changes.  The key here is to keep the second resistor consistent, then you can use the voltage between the sensor to represent the value, as voltage, of whatever you are trying to read.

Something else you may try is measuring the resistance between the input of the guage and the 12vdc pin.  If you can figure out the resistance of the second resistor (assuming they are using a voltage divider) then you can actually calculate the voltage at any point, and then set up a comparator circuit to work.  This would have to assume the gauge is operating at 12vdc and not 5vdc and is probably a long shot because of this.  For accuracy and repeatability you would want to feed the sensor with a stable voltage, so I wouldn't be surprised if it did have a 5vdc regulator driving it.

Speaking of comparator circuits, I just built one yesterday using an LM324 from Radioshack.  Pretty straight forward little circuits.

Anyway, back to your situation - there are two reasons I strongly believe that you are working with a voltage divider 1. that is the typical way to interface a resistive sensor and 2. because your sensor goes to 0 ohms.  Almost any other type of interface would be damaged by the sensor essentially shorting to ground.  However, if there is a second resistor in series with the 0 ohm sensor then the power supply won't actually be shorted.