### Theory:

Let us look at the following graph:

Let the mid-point be $$M$$ with the vertex $$M(x$$, $$y)$$.

From the similarity property, it is understood that $$\triangle{AMM'}$$ and $$\triangle{MDB}$$ are similar.

This makes the two corresponding sides equal.

Therefore, the ratio between the two corresponding sides are also equal.

$$\frac{AM'}{MD} = \frac{MM'}{BD} = \frac{AM}{MB} = \frac{1}{1}$$

We know that since $$M$$ is the mid-point of $$AB$$, $$AM = MB$$.

So, $$\frac{x - x_1}{x_2 - x} = \frac{y - y_1}{y_2 - y} = \frac{1}{1}$$

Let us only consider $$\frac{x - x_1}{x_2 - x} = \frac{1}{1}$$

$$x - x_1 = x_2 - x$$

$$2x = x_1 + x_2$$

$$x = \frac{x_1 + x_2}{2}$$

Similarly $$y$$ becomes $$\frac{y_1 + y_2}{2}$$

$$y = \frac{y_1 + y_2}{2}$$

Hence, the mid-point be $$M$$ with the vertex $$M(x$$, $$y)$$ becomes $$M(\frac{x_1 + x_2}{2}$$, $$\frac{y_1 + y_2}{2})$$.