a x + b y + c = 0 : Why It Applies to All Straight Lines

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: y - y_1 = k (x-x_1) where (x_1, y_1) is a point on the line, and k 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: y = k x + b where k is the slope, b 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: \frac{y-y1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} where (x_1, y_1), (x_2, y_2) are two points on the line. However, this form can represent neither line perpendicular nor parallel to x-axis due to the fact when x_1 = x_2 or y_1 = y_2, the form breaks down from dividing by zero.

[4] Point-Intercept form: \frac{x}{a} + \frac{y}{b} = 1 where a, b are the intersects the line made on x-axis and y-axis respectively, and a\neq 0, b\neq0. 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: a x +b y +c = 0 (a^2+b^2 \neq 0), 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 (x_1, y_1), (x_2, y_2), and any point (x, y) on the line connecting (x_1, y_1) and (x_2, y_2), the area of triangle with vertices (x_1, y_1), (x_2, y_2) and (x, y) must be zero!

Recall a theorem proved in my blog “Had Heron Known Analytic Geometry“, it means for such (x_1, y_1),  (x_2, y_2) and (x, y),

\left|\begin{array}{ccc}  x_1 & y_1 & 1  \\  x_2 & y_2 & 1 \\ x & y  & 1  \end{array}\right|= 0.

Therefore, we can define the line connecting two distinct points as a set of (x, y) such that the area of the triangle with vertices (x_1, y_1), (x_2, y_2) and (x, y) is zero, mathematically written as

A \triangleq  \{ (x, y)  | \left|\begin{array}{ccc}  x_1 & y_1 & 1  \\  x_2 & y_2 & 1 \\ x & y  & 1  \end{array}\right|= 0, (x_1-x_2)^2+(y_1-y_2)^2 \neq 0\}.

Since \forall (x, y) \in A,

\left|\begin{array}{ccc}  x_1 & y_1 & 1  \\  x_2 & y_2 & 1 \\ x & y  & 1  \end{array}\right|= x_1 y_2-x y_2-x_2 y_1+x y_1+x_2 y-x_1y =

(y-y_1)(x_1-x_2)-(x-x_1)(y_1-y_2)=0\quad\quad\quad\quad(1)

is an algebraic representation of the line connecting two distinct points (x_1, y_1) and (x_2, y_2).

When x_1=x_2, (1) becomes

(x-x_1)(y_1-y_2)=0,

and when x_1 = x_2, y_1-y_2 \neq  0, we have

x = x_1,

a line perpendicular to the horizontal axis.

When y_1=y_2, (1) becomes

y = y_1,

a line parallel to the horizontal axis.

Evaluate (1) with x_2=0, y_2=0 yields:

(y-y_1) x_1 -(x-x_1)y_1=0.

Collecting terms in (1), and letting

a=y_1-y_2,

b=x_2-x_1,

c=x_1y_2-x_2y_1,

(1) can be expressed as

ax + by + c = 0.

In fact, we can prove the following theorem:

B \triangleq  \{ (x, y) | \exists a, b, a^2+b^2 \neq 0, a x +b y+c=0\} \implies A=B.

To prove A=B, we need to show

\forall (x, y) \in A \implies (x, y) \in B\quad\quad\quad\quad(2)

\forall (x, y) \in B \implies (x, y) \in A\quad\quad\quad\quad(3)

We have already shown (2) by setting the values of a, b and c earlier.

We will prove (3) now:

\forall (x_1, y_1), (x_2, y_2) and (x, y) \in B, we have

\begin{cases}a x_1 + b y_1 +c =0 \\ a x_2 + b y_2 +c =0 \\ a x + b y+c =0\end{cases}.

Written in matrix form,

\left(\begin{array}{ccc}  x_1 & y_1 & 1  \\  x_2 & y_2 & 1 \\ x & y  & 1  \end{array}\right) \left(\begin{array}{rrr}  a \\  b \\  c  \end{array}\right)= 0.

If

\left|\begin{array}{ccc}  x_1 & y_1 & 1  \\  x_2 & y_2 & 1 \\ x & y  & 1  \end{array}\right| \neq 0,

then by Cramer’s rule,

\left(\begin{array}{rrr}  a \\  b \\  c  \end{array}\right) is a column vector of zeros,

i.e.,

a=b=c=0

which contradicts the fact that

a,b are not all zero.

Hence,

\left|\begin{array}{ccc}  x_1 & y_1 & 1  \\  x_2 & y_2 & 1 \\ x & y  & 1  \end{array}\right| = 0

which implies:

(x, y) \in A.

The consequence of A=B is that every point (x, y) on a line connecting two distinct points satisfies equation a x + b y + c =0 for some a, b (a^2+b^2\neq 0).

Stated differently,

a x + b y +c = 0 where a, b are not all zero is the governing equation of any straight line.

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