Theory:

Until now, we have only dealt with mid-point and the points of trisection.
Mid-point divides the line segment into two halves and the points of trisection divides the line segment into $$3$$ equal parts.
But, is it possible to divide the line segment into two unequal parts?

Yes, a line segment can be divided into two unequal parts using the section formula.

Imagine you have $$6$$ milk packets and two bags of unequal sizes.

Bag $$A$$ can hold $$4$$ milk packets while bag $$B$$ can hold only $$2$$ milk packets.

In this case, a total of $$6$$ milk packets is distributed across the two bags in the ratio of $$4:2$$.

Similarly, a line segment can also be divided in unequal ratios.
Let us look at how a section formula gets constructed.

In the figure given above, a line segment $$AB$$ is divided into two unequal parts in the ratio $$m : n$$.

Let $$A$$ be $$x_1$$, $$P$$ be $$x$$ and $$B$$ be $$x_2$$ such that $$x_2 > x > x_1$$.

The co-ordinate of $$P$$ divides the line segment in the ratio $$m : n$$.

This means, $$\frac{AP}{PB} = \frac{m}{n}$$

$$\frac{x - x_1}{x_2 - x} = \frac{m}{n}$$

$$m(x_2 - x)$$ $$=$$ $$n(x - x_1)$$

$$mx_2 - mx = nx - nx_1$$

$$mx_2 + nx_1 = mx + nx$$

$$x$$ $$=$$ $$\frac{mx_2 + nx_1}{m + n}$$

If $$A$$, $$P$$, and $$B$$ has the co-ordinates $$(x_1$$, $$y_1)$$, $$(x$$, $$y)$$, and $$(x_2$$, $$y_2)$$ respectively, then:

$$x$$ $$=$$ $$\frac{mx_2 + nx_1}{m + n}$$

$$y$$ $$=$$ $$\frac{my_2 + ny_1}{m + n}$$