1. **Problem Statement:** Given a line with intercepts $a$ and $b$ on the coordinate axes, and $p$ as the length of the perpendicular from the origin to this line, show that $$\frac{a}{l} + \frac{b}{1} = \frac{d}{l}.$$ 2. **Understanding the line equation:** The line with intercepts $a$ on the x-axis and $b$ on the y-axis can be written as $$\frac{x}{a} + \frac{y}{b} = 1.$$ 3. **Length of perpendicular from origin to the line:** The general form of the line is $$\frac{x}{a} + \frac{y}{b} = 1 \implies \frac{b x + a y}{a b} = 1 \implies b x + a y - a b = 0.$$ 4. **Formula for perpendicular distance from origin $(0,0)$ to line $Ax + By + C = 0$ is:** $$p = \frac{|A \cdot 0 + B \cdot 0 + C|}{\sqrt{A^2 + B^2}} = \frac{|C|}{\sqrt{A^2 + B^2}}.$$ 5. **Apply to our line:** Here, $A = b$, $B = a$, and $C = -a b$, so $$p = \frac{| -a b |}{\sqrt{b^2 + a^2}} = \frac{a b}{\sqrt{a^2 + b^2}}.$$ 6. **Interpreting the given expression:** The problem states $$\frac{a}{l} + \frac{b}{1} = \frac{d}{l}.$$ This seems to be a typographical or symbolic expression needing clarification. Assuming $l$ and $d$ relate to the line parameters or distances, the standard relation is the perpendicular distance $p$ as above. 7. **Summary:** The length of the perpendicular from the origin to the line with intercepts $a$ and $b$ is $$p = \frac{a b}{\sqrt{a^2 + b^2}}.$$ If $l$ and $d$ are defined differently, please provide their definitions for precise proof.