5. Intercept Form

Method I

We already know one method of finding the equation of a line, given its slope and y-intercept. The y-intercept is b. So the equation must look like y = mx + b.

What about m? Recall that m equals tanθ, where θ is the angle that the line makes with the X axis. Can we relate tanθ to a and b?

Here’s the figure again.

In triangle OAB, we have:

tan(π-θ) = OB/OA = b/a

⇒ -tanθ = b/a

⇒ m = -b/a

And we’re done. The required equation is

y = -(b/a)x+b,

This can be rearranged to get a better looking equation

x/a + y/b = 1

This is the intercept form of the equation of a line. Let’s look at the next method.

Method II

The previous method used the slope-intercept form of a line’s equation. Suppose we didn’t know about that form.

Then, to find the equation, we’ll take a random point P(x, y) on the line, and find a relation between its coordinates that always holds true.

Let’s join P to O.

Then, we can express the area of ΔOAB in terms of the areas of ΔPOA and ΔPOB.

We have:

Ar(OAB) = Ar(OAP) + Ar(OPB)

That is,

1/2 . a . b = 1/2 . a . y + 1/2 . b . x

After rearranging the terms, we get the same equation as above:

x/a + y/b = 1

Here’s a simulation where you can observe the intercept form of the line. Try changing the intercepts and see how the line changes.

Lesson Summary

1. The equation of the line, which makes intercepts of a and b on the X and Y axis respectively, is given by x/a + y/b = 1.
2. The points of intersection of this line with the X and Y axis are (a, 0) and (0, b) respectively.

Try deriving the equation by taking P in the 2nd quadrant. You should get the same equation. Take care of the signs of a, b, x and y.

The next lesson will cover a few examples to illustrate the intercept form of the line. See you there!