Transform the equation $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$ into the normal form when a > 0 and b > 0. If the perpendicular distance of straight line from the origin is p, deduce that $\frac{1}{{\mathrm{p}}^2}=\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}$

$\text{ Given line is }\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1(\mathrm{a}>0,\mathrm{b}>0\Rightarrow \mathrm{ab}>0)$
$\Rightarrow \mathrm{bx}+\mathrm{ay}=\mathrm{ab}$
Divide on both sides with $\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$
$\Rightarrow \frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{x}+\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{y}=\frac{\mathrm{ab}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$
Which is the normal form
Perpendicular distance from origin to the plane is
$\mathrm{p}=\frac{\mathrm{ab}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\Rightarrow \frac{1}{\mathrm{p}}=\frac{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}{\mathrm{ab}}$
S.O.B.S.
$$\begin{gathered} \Rightarrow \frac{1}{{\mathrm{p}}^2}=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2{\mathrm{b}}^2} \\ \therefore \frac{1}{{\mathrm{p}}^2}=\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2} \end{gathered}$$