In the traditional teaching of Analytical Geometry, the governing equation for a straight line has the following five forms, along with limitations for the first four:

[1] Point-Slope form: where is a point on the line, and is the slope. The limitation for this form is that it can not represent line perpendicular to the x-axis since it has no slope.

[2] Slope-Intercept form: where is the slope, is the intersect the line made on y-axis. Its limitation is that it can not represent line perpendicular to the x-axis.

[3] Two-Point form: where are two points on the line. However, this form can represent neither line perpendicular nor parallel to x-axis due to the fact when or , the form breaks down from dividing by zero.

[4] Point-Intercept form: where are the intersects the line made on x-axis and y-axis respectively, and . Again, this form can represent neither line perpenticular nor parallel to the x-axis. It does not work for any line that passes the point of origin either.

[5] General form: , this form can represent all lines.

Here I am presenting a proof to show [5] is indeed capable of representing all straight lines.

Let us start with an observation:

In a rectangular coordinate system, given two distinct points , and any point on the line connecting and , the area of triangle with vertices and must be zero!

Recall a theorem proved in my blog “Had Heron Known Analytic Geometry“, it means for such and ,

.

Therefore, we can define the line connecting two distinct points as a set of such that the area of the triangle with vertices and is zero, mathematically written as

.

Since ,

is an algebraic representation of the line connecting two distinct points and .

When , (1) becomes

,

and when , we have

,

a line* perpendicular* to the horizontal axis.

When , (1) becomes

,

a line *parallel* to the horizontal axis.

Evaluate (1) with yields:

.

Collecting terms in (1), and letting

,

,

,

(1) can be expressed as

.

In fact, we can prove the following theorem:

.

To prove , we need to show

We have already shown (2) by setting the values of and earlier.

We will prove (3) now:

and , we have

.

Written in matrix form,

.

If

,

then by Cramer’s rule,

is a column vector of zeros,

i.e.,

which contradicts the fact that

are not all zero.

Hence,

which implies:

.

The consequence of is that every point on a line connecting two distinct points satisfies equation for some .

Stated differently,

where are not all zero is the governing equation of any straight line.