# Intercept Form

## Intercept Form

This lesson will be focused on the intercept form of the straight line. I’ll illustrate two different methods of deriving the equation. (there may be more)

In this case, the line makes intercepts a and b with the X and Y axis respectively. What does that mean? Look at the figure below..

Note that both a and b can be positive, negative or zero, which tells the position of the points, at which the line cuts the axes, relative to the origin. The distance of the points from the origin will be |a| and |b|.

Lets find the equation then..

### Method I

We already know one method of finding the equation of a line, provided its slope and y-intercept are given. We’ve been given the y-intercept as b. So the equation must look like y = mx + b.

What about m? If only we could find it.. can we? Of course we can !

Recall that m is the tangent of the angle that the line makes with the X axis. Our aim is to express that in terms of a and b.

In triangle OAB, we have tan(π-θ)=OB/OA=b/a

Now, tan(π-θ) equals -tanθ which is -m. Therefore, -m=b/a, or m=-b/a. And we’re done !

The required equation is y=-(b/a)x+b, which can be rearranged to get a better looking equation, x/a+y/b=1

Here’s another method..

### Method II

The previous method assumed knowledge of the slope intercept form of a line’s equation.

Suppose we had no idea about the slope intercept form. Then we’ll get back to the same old method of taking a point, and establishing a relation between its coordinates (according to the given information).

Constructions: O is joined to P, perpendiculars PC and PD to the axes.

Now, area of the triangle OAB can always be expressed in the terms of the areas of triangles PBO and POA.

In this case we have, Ar(OAB)=Ar(OAP)+Ar(OPB)

That is, 1/2OAxOB=1/2OAxPC+1/2OBxPC, or 1/2ab=1/2ay+1/2bx

After getting rid of the 1/2 and rearranging, we get the same equation as above: x/a+y/b=1

By the way the coordinates of A and B will be (a, 0) and (0, b) respectively.

Why? Because, the y-coordinate of every point (one of which is A) on the X-axis would be 0 (The equation of the X axis is y=0). Substituting 0 for y in the above equation we get x = a. Hence the coordinates (a, 0). Same is the story for B’s coordinates.

Remember these two, as they will be quite helpful in solving problems.

## Lesson Summary

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

As an exercise, 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 !