If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b,

Solution:

It is known that the equation of a line whose intercepts on the axes are a and b is

$\frac{x}{a}+\frac{y}{b}=1$

or $b x+a y=a b$

or $b x+a y-a b=0$ $\ldots(1)$

The perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $\left(x_{1}, y_{1}\right)$ is given by $d=\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}$.

On comparing equation (1) to the general equation of line $A x+B y+C=0$, we obtain $A=b, B=a$, and $C=-a b$.

Therefore, if $p$ is the length of the perpendicular from point $\left(x_{1}, y_{1}\right)=(0,0)$ to line $(1)$, we obtain

$p=\frac{|A(0)+B(0)-a b|}{\sqrt{b^{2}+a^{2}}}$

$\Rightarrow p=\frac{|-a b|}{\sqrt{a^{2}+b^{2}}}$

On squaring both sides, we obtain

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

$\Rightarrow p^{2}\left(a^{2}+b^{2}\right)=a^{2} b^{2}$

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

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

Hence, we showed that $\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$.