The Intercepts Form of a Line

If a and b are non-zero X and Y intercepts of a line l, then its equation is of the form

\frac{x}{a} + \frac{y}{b} = 1

Since a is an X-intercept of the line l, and as we know that if any point lies on the X-axis its value of Y is equal to zero, it passes through the point A\left( {a,0} \right). Also if b is the Y-intercept of the line l, and we know that any point that lies on the Y-axis has a value of X equal to zero, it passes through the point B\left( {0,b} \right) as shown in the given diagram.


inercepts-form-line

Now to prove the intercepts form of a line, use the formula for the two points form of a straight line as given by

\frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{x - {x_1}}}{{{x_2} - {x_1}}}

Take A\left( {a,0} \right) = \left( {{x_1},{y_1}} \right) and B\left( {0,b} \right) = \left( {{x_2},{y_2}} \right), and put these values in the above formula:

\begin{gathered} \Rightarrow \frac{{y - 0}}{{b - 0}} = \frac{{x - a}}{{0 - a}} \\ \Rightarrow \frac{y}{b} = \frac{x}{{ - a}} - \frac{a}{{ - a}} \\ \end{gathered}

 \Rightarrow \boxed{\frac{x}{a} + \frac{y}{b} = 1}

This is the required equation of a straight line in intercepts form.

Example: Find the equation of a straight line with X-intercept A\left( {3,0} \right) and Y-intercept B\left( {0,2} \right).

From the above information we know the X-intercept is a = 3 and the Y-intercept is b = 2. Now we put all these values in the formula of the intercepts form as given:

\begin{gathered} \frac{x}{a} + \frac{y}{b} = 1 \\ \Rightarrow \frac{x}{3} + \frac{y}{2} = 1 \\ \Rightarrow 2x + 3y = 6 \\ \Rightarrow 2x + 3y - 6 = 0 \\ \end{gathered}

This is the required equation of a straight line.