$\dfrac{x}{a}+\dfrac{y}{b} \,=\, 1$

It is an equation of a straight line when a straight line intersects both axes with $x$ and $y$ intercepts.

A straight line intersects both horizontal and vertical axes of Cartesian coordinate system at $P$ and $Q$ points with $a$ as $x$-intercept and $b$ as $y$-intercept.

The intersecting points of straight line $\small \overleftrightarrow{PQ}$ in coordinate form are $P(a, 0)$ and $Q(0, b)$.

$A(x, y)$ is any point on the straight line $\small \overleftrightarrow{PQ}$. Draw a perpendicular line to $x$-axis from point $A$ and it intersects the $x$-axis at point $B$.

Similarly, draw a perpendicular line to $y$-axis from point $A$ and it intersects the $y$-axis at point $C$.

It formed two right triangles $\Delta APB$ and $\Delta QAC$. The adjacent side of $\Delta APB$ is on $x$-axis and the adjacent side of $\Delta QAC$ is parallel to the $x$-axis. Similarly, the opposite side of $\Delta QAC$ is on $y$-axis and the opposite side of $\Delta APB$ is parallel to the $y$-axis. The hypotenuse of both right triangles are part of straight line $\small \overleftrightarrow{PQ}$.

Therefore, the right triangles $\Delta APB$ and $\Delta QAC$ are similar triangles. Hence, the corresponding angles of both triangles are same.

If $\angle QAC$ is equal to $\alpha$, then $\angle APB$ is also equal to $\alpha$.

According to $\Delta APB$

$\tan{\alpha} \,=\, \dfrac{AB}{BP}$

$\implies \tan{\alpha} \,=\, \dfrac{AB}{OP-OB}$

$\,\,\, \therefore \,\,\,\,\,\, \tan{\alpha} \,=\, \dfrac{y}{a-x}$

According to $\Delta QAC$

$\tan{\alpha} \,=\, \dfrac{QC}{AC}$

$\implies \tan{\alpha} \,=\, \dfrac{OQ-OC}{AC}$

$\,\,\, \therefore \,\,\,\,\,\, \tan{\alpha} \,=\, \dfrac{b-y}{x}$

According to both $\Delta APB$ and $\Delta QAC$

$\tan{\alpha} \,=\, \dfrac{y}{a-x} \,=\, \dfrac{b-y}{x}$

Now, simplify the equation to get the equation of a straight line in intercept form.

$\dfrac{y}{a-x} \,=\, \dfrac{b-y}{x}$

$\implies xy \,=\, (a-x)(b-y)$

$\implies xy \,=\, a(b-y)-x(b-y)$

$\implies xy \,=\, ab-ay-xb+xy$

$\implies \require{cancel} \cancel{xy} \,=\, ab-ay-xb+\cancel{xy}$

$\implies 0 \,=\, ab-ay-xb$

$\implies bx+ay \,=\, ab$

$\implies \dfrac{bx+ay}{ab} \,=\, 1$

$\implies \dfrac{bx}{ab}+\dfrac{ay}{ab} \,=\, 1$

$\implies \require{cancel} \dfrac{\cancel{b}x}{a\cancel{b}}+\dfrac{\cancel{a}y}{\cancel{a}b} \,=\, 1$

$\,\,\, \therefore \,\,\,\,\,\, \dfrac{x}{a}+\dfrac{y}{b} \,=\, 1$

It is a linear equation in algebraic form and represents an equation of straight line when a straight line intersects both axes with double intercept.

Latest Math Topics

Latest Math Problems

Email subscription

Math Doubts is a free math tutor for helping students to learn mathematics online from basics to advanced scientific level for teachers to improve their teaching skill and for researchers to share their research projects.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.