If $‘p’$  is the length of the perpendicular from origin to the line $\frac{x}{a}+\frac{y}{b}=1$ then

If $\text{'}p\text{'}$  is the length of perpendicular from origin to the line $\frac{x}{a}+\frac{y}{b}=1$

$\left(0,0\right),bx+ay-ab=0$

$p=\left|\frac{-ab}{\sqrt{b^2+a^2}}\right|$

$p=\frac{ab}{\sqrt{a^2+b^2}}$

$p^2=\frac{a^2{b}^2}{a^2+b^2}$

$a^2+b^2=\frac{a^2{b}^2}{p^2}$

$\frac{a^2+b^2}{a^2{b}^2}=\frac{1}{p^2}$

$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}$