(i) Since,  A  lies on the  x-axis, let the coordinates of  A  be  (x, 0).

Since  B  lies on the  y - axis, let the coordinates of  B  be  (0, y).

Let  m = 1   and    n = 2.

Using section formula, 

$$\begin{gathered} \mathrm{Co}-\mathrm{ordinates} \mathrm{of} \mathrm{P} = \left( \frac{1 ( 0 ) + 2 ( \mathrm{x} )}{1 + 2}, \frac{1 \mathrm{y} + 2 ( 0 )}{1 + 2} \right) \\ \Rightarrow ( 4, - 1 ) = \left( \frac{2 \mathrm{x}}{3}, \frac{\mathrm{y}}{3} \right) \\ \Rightarrow \frac{2 \mathrm{x}}{3} = 4 \mathrm{and} \frac{\mathrm{y}}{3} = - 1 \\ \Rightarrow 2 \mathrm{x} = 3 \mathrm{x} 4 \mathrm{and} \mathrm{y} = - 1 \mathrm{x} 3 \\ \Rightarrow \mathrm{x} = \frac{12}{2} \mathrm{and} \mathrm{y} = - 3 \\ \Rightarrow \mathrm{x} = 6 \mathrm{and} \mathrm{y} = - 3 \end{gathered}$$

So, the co-ordinates of  A  are ( 6, 0 )  and that of  B  are  (0, - 3 ).

$$\begin{gathered} ( \mathrm{ii} ) \mathrm{Slope} \mathrm{of} \mathrm{AB} = \frac{- 3 - 0}{0 - 6} = \frac{- 3}{- 6} = \frac{1}{2} \\ \Rightarrow \mathrm{Slope} \mathrm{of} \mathrm{line} \mathrm{perpendicular} \mathrm{to} \mathrm{AB} = \mathrm{m} =- 2 \\ \mathrm{P} ( 4, - 1 ) \\ \Rightarrow \mathrm{Required} \mathrm{equation} \mathrm{is} \\ \mathrm{y} - {\mathrm{y}}_1 = \mathrm{m} ( \mathrm{x} - {\mathrm{x}}_1 ) \\ \Rightarrow \mathrm{y} - ( - 1 ) = - 2 ( \mathrm{x} - 4 ) \\ \Rightarrow \mathrm{y} + 1 = - 2 \mathrm{x} + 8 \\ \Rightarrow 2 \mathrm{x} + \mathrm{y} = 7. \end{gathered}$$