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.
 
 
fig_1.svg
 
 
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:
 
\(x\) \(=\) \(\frac{mx_2 + nx_1}{m + n}\)
 
\(y\) \(=\) \(\frac{my_2 + ny_1}{m + n}\)