A and B are points on the positive x-axis and y -axis respectively. The point P(4, 3) divides the line segment joining AB in the ratio 3: 1. Find the equation of the line AB.

We know that, the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is:

 $\mathrm{y}-{\mathrm{y}}_1=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}\left(\mathrm{x}-{\mathrm{x}}_1\right)$ 
Given, A and B are points on the positive x-axis and y-axis. Hence let us assume that A and B have the coordinates of (x, 0) and (0, y), respectively.
Also, P divides AB in the ratio  . 

Therefore, by using the section formula,

$\mathrm{P}(\mathrm{x},\mathrm{y})=\left(\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}},\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}}\right)$ ...(i)
$\begin{array}{l}=\left(\frac{3\times 0+1\times \mathrm{x}}{3+1},\frac{3\times \mathrm{y}+1\times 0}{3+1}\right)\\ =\left(\frac{\mathrm{x}}{4},\frac{3\mathrm{y}}{4}\right)\end{array}$ 
But the given coordinates of P are P(4, 3). 
So, 

$\Rightarrow \frac{\mathrm{x}}{4}=4\text{ and }\frac{3\mathrm{y}}{4}=3$

$\Rightarrow \mathrm{x}=16\text{ and }\mathrm{y}=4$

As a result, the coordinates of point A are (16, 0) while the coordinates of point B are (0, 4).
Now from (i) the equation of line AB is,

$\mathrm{y}-0=\frac{4-0}{0-16}(\mathrm{x}-16)$

$\Rightarrow \mathrm{y}=-\frac{1}{4}(\mathrm{x}-16)$

$\Rightarrow 4\mathrm{y} + \mathrm{x}=16$ 
Hence, the equation of line AB is x + 4y = 16.