Equation of Straight Line in Plane/Two-Intercept Form/Proof 1

Theorem

Let $\LL$ be a straight line which intercepts the $x$-axis and $y$-axis respectively at $\tuple {a, 0}$ and $\tuple {0, b}$, where $a b \ne 0$.

Then $\LL$ can be described by the equation:

$\dfrac x a + \dfrac y b = 1$

From the General Equation of Straight Line in Plane, $\LL$ can be expressed in the form:

$(1): \quad \alpha_1 x + \alpha_2 y = \beta$

where $\alpha_1, \alpha_2, \beta \in \R$ are given, and not both $\alpha_1, \alpha_2$ are zero.

Substituting for the two points whose coordinates we know about:

$\ds x = a, y = 0: \, $

$\ds \alpha_1 \times a + \alpha_2 \times 0$

$=$

$\ds \beta$

$\ds \leadsto \ \ $

$\ds \alpha_1$

$=$

$\ds \dfrac \beta a$

and:

$\ds x = 0, y = b: \, $

$\ds \alpha_1 \times 0 + \alpha_2 \times b$

$=$

$\ds \beta$

$\ds \leadsto \ \ $

$\ds \alpha_2$

$=$

$\ds \dfrac \beta b$

We know that $\beta \ne 0$ because none of $a, b, \alpha_1, \alpha_2$ are equal to $0$.

Hence:

$\ds \dfrac \beta a x + \dfrac \beta b y$

$=$

$\ds \beta$

substituting for $\alpha_1$ and $\alpha_2$ in $(1)$

$\ds \leadsto \ \ $

$\ds \dfrac x a + \dfrac y b$

$=$

$\ds 1$

dividing both sides by $\beta$

$\blacksquare$