### Theory:

Let us find the equation of the straight line whose intercepts are $$a$$ and $$b$$ on the coordinate axes, respectively.

Let $$PQ$$ be a straight line that meets the $$x$$ - axis at the point $$A$$ and the $$y$$ axis at the point $$B$$. The coordinates of points $$A$$ and $$B$$ are $$(a,0)$$ and $$(0,b)$$, respectively. Then, $$OA = a$$ and $$OB = b$$.

Here, $$(x_1,y_1) = (a,0)$$ and $$(x_2,y_2) = (0,b)$$.

Substituting the known values in two point form equation, we get:

$$\frac{y - 0}{b - 0} = \frac{x - a}{0 - a}$$

$$\frac{y}{b} = \frac{x - a}{-a}$$

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

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

Hence, the equation of the intercept of the line is $$\frac{x}{a} + \frac{y}{b} = 1$$.
Example:
Find the intercepts of the line $$-3x + 4y + 12 = 0$$ on the coordinate axes.

Solution:

The given equation is $$-3x + 4y + 12 = 0$$.

Let us write the equation in the normal form.

$$-3x + 4y = -12$$

$$\frac{-3}{-12}x + \frac{4}{-12}y = \frac{-12}{-12}$$ [Dividing by $$-12$$ on both sides so that the RHS is equal to $$1$$]

$$\frac{x}{4} + \frac{y}{-3} = 1$$

Comparing the above equation with the equation of the intercept of the line $$\frac{x}{a} + \frac{y}{b} = 1$$, we get:

$$a = 4$$ and $$b = -3$$

Therefore, the intercept of the line is $$a = 4$$ and $$b = -3$$.