1. **Problem Statement:** Given a line with intercepts $a$ and $b$ on the x-axis and y-axis respectively, and $p$ as the length of the perpendicular from the origin to this line, show that $$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}.$$ 2. **Equation of the line:** The line with intercepts $a$ and $b$ on the axes can be written as: $$\frac{x}{a} + \frac{y}{b} = 1.$$ 3. **Rewrite the line in standard form:** Multiply both sides by $ab$: $$b x + a y = ab.$$ 4. **Formula for perpendicular distance from origin to line:** The distance $p$ from the origin $(0,0)$ to the line $Ax + By + C = 0$ is: $$p = \frac{|A \cdot 0 + B \cdot 0 + C|}{\sqrt{A^2 + B^2}} = \frac{|C|}{\sqrt{A^2 + B^2}}.$$ Rewrite the line as: $$b x + a y - ab = 0,$$ so $A = b$, $B = a$, and $C = -ab$. 5. **Calculate $p$:** $$p = \frac{| -ab |}{\sqrt{b^2 + a^2}} = \frac{ab}{\sqrt{a^2 + b^2}}.$$ 6. **Find $\frac{1}{p^2}$:** $$\frac{1}{p^2} = \frac{1}{\left(\frac{ab}{\sqrt{a^2 + b^2}}\right)^2} = \frac{1}{\frac{a^2 b^2}{a^2 + b^2}} = \frac{a^2 + b^2}{a^2 b^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} = \frac{1}{b^2} + \frac{1}{a^2}.$$ 7. **Conclusion:** We have shown that: $$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2},$$ which is the required result. This completes the proof.