Constellation diagrams exist in what is called signal space which is an abstraction used to describe finite-energy signals. The coordinate axes, even if they are marked $x$ and $y$ as in Marcus Muller's answer, really represent unit-energy signals such as $$s_I(t) = \sqrt{\frac 2T}\cos(2\pi f_c t)\mathbf 1_{t \in [0,T)}(t),$$ $$s_Q(t) = -\sqrt{\frac 2T}\sin(2\pi f_c t)\mathbf 1_{t \in [0,T)}(t)$$ where $T = \frac{n}{f_c}$ and $\mathbf 1_{t \in [0,T)}(t)$ means a rectangular pulse of duration $T$. Note that both sinusoidal pulses consist of and integer number $n$ periods of a sinusoid of frequency $f_c$. Verify for yourself that $s_I(t)$ and $s_Q(t)$ are indeed unit energy signals and that they are orthogonal, that is, verify that
$$\int_0^T [s_I(t)]^2 \mathrm dt = \int_0^T [s_Q(t)]^2 \mathrm dt = 1;\quad
\int_0^T s_I(t)s_Q(t) \mathrm dt =0.$$
If we use $\pm As_I(t)$ as our PSK signals, then the signal energy is $A^2$(why?) and so if we use $E_b$ to denote the energy per bit, then
with respect to the signal space with axes $s_I(t)$ and $s_Q(t)$, the two signals used in PSK can be represented by the constellation points $(\sqrt{E_b},0)$ and $(-\sqrt{E_b},0)$. Left-handed folks might prefer to use PSK signals $\pm As_Q(t)$ in which case the signal constellation would have points $(0,\sqrt{E_b})$ and $(0,-\sqrt{E_b})$.
Mealy-mouthed ambidextrous people might want to use
$$\pm A[s_I(t)\cos(\theta)+s_Q(t)\sin(\theta)] = \pm A\sqrt{\frac 2T}\cos(2\pi f_c t+\theta)\mathbf 1_{t \in [0,T)}(t)$$
as PSK signals in which case in the signal space with axes $s_I(t)$ and $s_Q(t)$, the constellation points would be $\pm (\sqrt{E_b}\cos\theta,\sqrt{E_b}\sin\theta)$. But if we had chosen axes $\sqrt{\frac 2T}\cos(2\pi f_c t+\theta)$ and
$-\sqrt{\frac 2T}\sin(2\pi f_c t+\theta)$ instead (this just corresponds to a rotation of axes by $\theta$ from the previous signal space), then the constellation points would once again be $(\sqrt{E_b},0)$ and $(-\sqrt{E_b},0)$.
Thus, in signal space in which the coordinate axes represent unit-energy orthogonal signals, the energy of the signal corresponding to a constellation point equals the squared distance from the origin. Note that in (antipodal) PSK considered above, this squared distance from the origin is $E_b$ for each of the two constellation points. For QPSK (which I have ignored for the most part but you can read about it in this answer of mine), each of the four constellation points are at squared distance $2E_b$ from the origin which makes sense in that $E_b$ is the energy per bit and a QPSK signal carries two bits.
Finally, with regard to the importance of distance as a concept, given two constellation points in signal space, the probability of error (in an AWGN channel with two-sided power spectral density $N_0/2$) is $Q\left(\frac{d}{\sqrt{2N_0}}\right)$ where $d$ is the distance between the two points. For the PSK examples considered above, $d = 2\sqrt{E_b}$ which gives the familiar result
$$P_e = Q\left(\frac{d}{\sqrt{2N_0}}\right) = Q\left(\frac{2\sqrt{E_b}}{\sqrt{2N_0}}\right) = Q\left(\sqrt{\frac{2E_b}{N_0}}\right)$$
