If three points $(h,0)$, $(a,b)$ and $(0,k)$ lie on a line, show that $\dfrac{a}{h}+\dfrac{b}{k}=1$.

**OR**

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

Question

If three points $(h,0)$, $(a,b)$ and $(0,k)$ lie on a line, show that $\dfrac{a}{h}+\dfrac{b}{k}=1$.

**OR**

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

Accepted Answer

>

> [!TLDR]
> Using the intercept form of a line, the collinearity of the three points directly gives $\dfrac{a}{h}+\dfrac{b}{k}=1$.
>
> The points $(h,0)$ and $(0,k)$ are the x-intercept and y-intercept of a line. The equation of a line with x-intercept $h$ and y-intercept $k$ (intercept form) is:
>
> $\dfrac{x}{h}+\dfrac{y}{k} = 1$
>
> Since the point $(a,b)$ also lies on this same line (given the three points are collinear), it must satisfy the line's equation. Substituting $x=a, y=b$:
>
> $\dfrac{a}{h}+\dfrac{b}{k} = 1$
> …

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

Q.If three points $(h,0)$ , $(a,b)$ and $(0,k)$ lie on a line, show that $\dfrac{a}{h}+\dfrac{b}{k}=1$ . OR 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 $\dfrac{1}{P^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}$ .

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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