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

**OR**

Find the equation of the line which passes through the point $(2, 2)$ and cuts off interce

Question

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

**OR**

Find the equation of the line which passes through the point $(2, 2)$ and cuts off intercepts on the axes whose sum is $9$.

Accepted Answer

>

> [!TLDR]
> Using the intercept form of the line and the perpendicular-distance-from-origin formula proves $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$.
>
> A line with intercepts $a$ and $b$ on the axes has the intercept-form equation:
>
> $\dfrac{x}{a} + \dfrac{y}{b} = 1$, i.e. $bx + ay - ab = 0$
>
> The perpendicular distance from the origin $(0,0)$ to a line $Ax+By+C=0$ is $\dfrac{|A(0)+B(0)+C|}{\sqrt{A^2+B^2}}$.
>
> Here $A=b$, $B=a$, $C=-ab$, so:
>
> $p = \dfrac{|-ab|}{\sqrt{a^2+b^2}} = \dfrac{ab}{\sqrt{a^2+b^2}}$
>
> Squaring both sides: $p^2 = \dfrac{a^2b^2}{a^2+b^2}$
> …

Quadrant	x-sign	y-sign	Example
I	$+$	$+$	$(2, 3)$
II	$-$	$+$	$(-1, 4)$
III	$-$	$-$	$(-3, -2)$
IV	$+$	$-$	$(5, -1)$

Concept understanding — Coordinate Geometry

Coordinate Geometry: Where Algebra Meets Geometry

The Big Idea

The Coordinate System

A Concrete Example

The Four Quadrants

Why This Matters

The Precise Statement

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