# General Form

## General Form

Now that I’ve covered all forms of the straight line equations (except one), we must look at the general form of the equation of the straight line.

That is to mean, what kind of an equation would represent a straight line? And what kind wouldn’t?

There can be all sorts of equations.. like x + 5y – 6 = 0, xy = 3 and x2 + y= 4. Which of these would represent a straight line? And which of them will not?

I claim that an equation of the form Ax + By + C = 0, will represent a straight line. Here A, B, C are arbitrary constants (A and B cannot be both 0), and x and y are variables (which are the coordinates of points on the line)

Here’s a proof..

Suppose B≠0, then after dividing the equation by B and rearranging the terms we get y = (-A/B) x + (-C/B).

By putting -A/B = m and -C/B = c, the above equation becomes y = mx + c. This looks familiar, doesn’t it?

This is a line of slope ‘m’, and y-intercept ‘c’, as we derived here. Therefore, the equation Ax + By + C = 0 represents a line with slope ‘-A/B’ and y-intercept ‘-C/B’.

In case B = 0, then the equation will become x = -C/A. If we put -C/A = a, then the equation becomes x = a, which represents a line parallel to the Y axis (explained here).

Similarly, the equation can be transformed into something like $$\frac{x}{(-C/A)} + \frac{y}{(-C/B)} = 1$$ which is again a line, which has intercepts ‘-C/A’ and ‘-C/B’ on the coordinate axes (explained here).

Therefore, the equation Ax + By + C = 0 represents a straight line.

But what if we hadn’t derived these equations previously?

I’ll give one more proof.

The idea is to prove that any three points taken on the curve represented by Ax + By + C = 0 are collinear. Because if they are, then that curve can be nothing else but a line !

Let P(x1,y1) and Q(x2,y2) and R(x3,y3) be any three points on the curve whose equation is Ax + By + C = 0.

Then the coordinates of these three points must satisfy the equation. We get the following three relations.

Ax1 + By1 + C = 0 (I), Ax2 + By2 + C = 0 (II), Ax3 + By3 + C = 0 (III)

Now, the aim is to prove that, if these relations hold true, then the points must be collinear (we have many methods to do that).

Subtracting (I) from (II), we get A(x2-x1) = B(y2-y1) or (y2-y1)/(x2-x1) = A/B.

Similarly, after subtracting (II) from (III), we get (y3-y2)/(x3-x2) = A/B.

We therefore have (y2-y1)/(x2-x1) = (y3-y2)/(x3-x2), which on rearranging gives x1(y2-y3)+x2(y3-y1)+x3(y1-y2)=0.

On dividing this by 2, we get 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)] = 0. This means that the area of the triangle PQR is 0, proving that the points are collinear. (We’ve done something like this previously)

And we’re done! We took 3 random points on the curve, and proved that those 3 points are collinear. This can only be possible if the curve is a line. Therefore, the equation Ax + By + C = 0 must represent a line.

Phew ! This might be a little too complex for you. Read it once again, and again once more. Things will get better.

## Lesson Summary

1. Any equation of the form Ax + By + C = 0 (i.e. a linear equation in one or two variables), will represent a straight line on the X-Y plane, where both A and B should not be zero.