Equation of Straight Line in Plane/Two-Intercept Form

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Theorem

Let $\mathcal L$ 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 $\mathcal L$ can be described by the equation:

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


Proof

Straight-line-double-intercept-form.png


From the General Equation of Straight Line in Plane, $\mathcal L$ 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:

\(\, \displaystyle x = a, y = 0: \, \) \(\displaystyle \alpha_1 \times a + \alpha_2 \times 0\) \(=\) \(\displaystyle \beta\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \alpha_1\) \(=\) \(\displaystyle \dfrac \beta a\)
\(\, \displaystyle x = 0, y = b: \, \) \(\displaystyle \alpha_1 \times 0 + \alpha_2 \times b\) \(=\) \(\displaystyle \beta\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \alpha_2\) \(=\) \(\displaystyle \dfrac \beta b\)

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


Hence:

\(\displaystyle \dfrac \beta a x + \dfrac \beta b y\) \(=\) \(\displaystyle \beta\) substituting for $\alpha_1$ and $\alpha_2$ in $(1)$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac x a + \dfrac y a\) \(=\) \(\displaystyle 1\) dividing both sides by $\beta$

$\blacksquare$


Also known as

This form of the Equation of Straight Line in Plane is also known as the intercept form.


Sources