If $p$ is the length of the perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$,then show that $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$.

(N/A) The equation of a line with intercepts $a$ and $b$ on the axes is given by $\frac{x}{a} + \frac{y}{b} = 1$.
This can be rewritten as $bx + ay = ab$,or $bx + ay - ab = 0$ ... $(1)$.
The perpendicular distance $p$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Comparing equation $(1)$ with the general form,we have $A = b$,$B = a$,and $C = -ab$.
Substituting these values,we get $p = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}} = \frac{|-ab|}{\sqrt{a^2 + b^2}}$.
Squaring both sides,we get $p^2 = \frac{a^2 b^2}{a^2 + b^2}$.
Taking the reciprocal,we get $\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2}$.
Thus,$\frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2}$,which is $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.