Cool air far enough and its vapour pressure meets the saturation curve. From that instant the air can hold no more, and every further degree of cooling puts water on the nearest cold surface. That temperature is the dew point, $T_d$.

Two roads lead to it, and this lesson walks both. From temperature and relative humidity, the Magnus form: $T_d = \dfrac{c\,\gamma}{b - \gamma}$, with $\gamma = \ln\dfrac{\mathrm{RH}}{100} + \dfrac{b\,T}{c + T}$ — where $T$ is the dry-bulb temperature in °C, $\mathrm{RH}$ is the relative humidity as a percentage, $\gamma$ (gamma) is a bare intermediate with no units and no physical meaning of its own, and $b = 17.62$, $c = 243.12\ ^\circ\mathrm{C}$ are the WMO's coefficients. From humidity ratio and pressure, the chain runs $W \rightarrow p_v \rightarrow T_d$: find the vapour pressure with $p_v = \dfrac{W p}{0.62198 + W}$, then ask the saturation curve which temperature owns that pressure.

Note what is not in either answer: the air's own temperature does not appear in the second road at all. The dew point depends only on how much water the air is carrying and the pressure it carries it at. Heat the air and the dew point does not move a hair — the same fact the humidity ratio makes, stated as a temperature instead of a mass.

This is the number that decides whether a chilled-water pipe needs insulation, whether a supply duct will drip into somebody's ceiling tiles, and whether the slab will be wet in the morning. Below 0 °C what you have named is really a FROST point — the vapour deposits as frost rather than condensing as dew, and saturation over ice follows a slightly different curve. Treat a sub-zero answer as approximate and say so.