The equation $$|z - i| = |z - 1|$$, $$i = \sqrt{-1}$$, represents:

Let us denote the unknown complex number by $$z$$ and write it in its rectangular (Cartesian) form: we put

$$z = x + iy,$$

where $$x$$ and $$y$$ are real numbers representing the point $$(x,y)$$ in the Argand plane.

The given condition is

$$|z - i| = |z - 1|.$$

First we translate each modulus into the distance formula. Remember that for any two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$, the distance between them is

$$\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}.$$

Here $$z - i$$ means the vector from the fixed point $$i$$ (which is the point $$(0,1)$$) to the variable point $$z = x + iy$$. So

$$|z - i| = \sqrt{(x - 0)^2 + (y - 1)^2}.$$

Similarly, $$z - 1$$ connects the fixed point $$1$$ (the point $$(1,0)$$) to $$z$$, giving

$$|z - 1| = \sqrt{(x - 1)^2 + (y - 0)^2}.$$

Because these two lengths are equal, we have the equation

$$\sqrt{(x - 0)^2 + (y - 1)^2} = \sqrt{(x - 1)^2 + (y - 0)^2}.$$

Now we square both sides to eliminate the square roots:

$$(x - 0)^2 + (y - 1)^2 = (x - 1)^2 + (y - 0)^2.$$

Writing out every term explicitly gives

$$x^2 + (y - 1)^2 = (x - 1)^2 + y^2.$$

Next we expand each squared binomial step by step:

$$x^2 + \bigl(y^2 - 2y + 1\bigr) = \bigl(x^2 - 2x + 1\bigr) + y^2.$$

Combining like terms on the left side we obtain

$$x^2 + y^2 - 2y + 1 = x^2 - 2x + 1 + y^2.$$

Notice that the terms $$x^2$$ on both sides cancel, and the terms $$y^2$$ on both sides also cancel, leaving

$$-\,2y + 1 = -\,2x + 1.$$

Subtract $$1$$ from each side; the ones disappear:

$$-\,2y = -\,2x.$$

Now divide every term by $$-2$$ to isolate $$y$$:

$$y = x.$$

Thus the locus of points satisfying the original equation is the straight line

$$y = x,$$

which plainly passes through the origin and has slope $$1$$.

Hence, the correct answer is Option C.