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 =


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


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




(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.


\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,



which contradicts the fact that

a,b are not all zero.


\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.


Had Heron Known Analytic Geometry…

In my previous two posts, “An Algebraic Proof of Heron’s Formula” and “An Alternative Derivation of Heron’s Formula,” I proved Heron’s formula for the area of a triangle with three given sides.  Based on Heron’s formula, we can now prove a theorem concerning the area of any triangle in a rectangle coordinate system, namely,

The area of a triangle with vertices at (x_1, y_1), (x_2, y_2), (x_3, y_3) in a rectangle coordinate system can be expressed as


where D is the determinant of matrix:

\left(\begin{array}{ccc}  x_1 & y_1 & 1  \\  x_2 & y_2 & 1 \\ x_3 & y_3  & 1  \end{array}\right)

I offer the following proof:

Screen Shot 2017-06-14 at 1.02.37 PM.png

Fig. 1

By Heron’s formula, the area of triangle in Fig. 1

A=\sqrt{s (s-a) (s-b) (s-c)}\quad\quad\quad\quad\quad(2)

where a, b, c are three sides of the triangle and, s=(a+b+c)/2.







Let B=|\frac{1}{2} D|,  we have

B^2 ={|\frac{1}{2} D|}^2=(\frac{1}{2}D)^2.

Compute A^2-B^2 using Omega CAS Explorer (see Fig. 2) , the result shows


Screen Shot 2017-06-13 at 9.21.21 PM.png

Fig. 2

Since A> 0, B\geq 0 implies

A+B >0,

A^2-B^2=(A-B)(A+B)=0 implies



A = B.

Hence, (1) and (2) are equivalent.

I would like to learn any other alternative proof.

Higham’s Parametric Curve

Screen Shot 2017-06-10 at 11.02.26 AM.png

Fig. 1

The parametric curve (see Fig. 1) has always intrigued me. It was to my delight to finally find its equations from “MATLAB guide” written by the Highams:

x = \int\limits_{0}^{t} \sin {\omega^2}\; d\omega

y = \int\limits_{0}^{t} \cos {\omega^2}\; d\omega

Below is the curve plotted by Omega CAS Explorer:

Screen Shot 2017-06-10 at 12.43.19 PM.png

Fig. 2

After several failed trials, I realized that ‘nticks’ must be provided in ‘plot2d’ in order to produced the image correctly.

Next, I tried ‘draw2d’ function (see Fig. 3), but the tics and numbers are too close to the image of the curve.

Screen Shot 2017-06-10 at 12.36.10 PM.png

Fig. 3

To better position the image,  I specified ‘xrange’ and ‘yrange’ to put more space between the image and the tics and numbers. Cropping the resulting image to obtain the Fig. 1 at the top of this post.

Screen Shot 2017-06-10 at 12.39.31 PM

Fig. 4

I would like to ask all the maxima Jedis out there,

Without specify ‘xrange’ and ‘yrange’, is there an option that I can set to turn off the tics and numbers ?

Thank you in advance for your answer.


An Alternative Derivation of Heron’s Formula

In my previous blog titled “An Algebraic Proof of Heron’s Formula“, I algebraically derived the Heron’s formula concerning A, the area of a triangle:


where a, b, c are three sides of a triangle and s=\frac{a+b+c}{2}.

There is an alternative derivation which requires some basic trigonometry.

Here it is:

Screen Shot 2017-06-02 at 9.46.32 PM.png

Fig. 1

Let \theta = \angle ABC (see Fig. 1),  then

A=\frac{1}{2}a\;h = \frac{1}{2}a\;(c\sin{\theta})


A^2=\frac{1}{4}a^2 c^2 \sin^2{\theta} = \frac{1}{4} a^2 c^2 (1-\cos^2{\theta})\quad\quad\quad\quad\quad(1)

due to trigonometric identity \sin^2{\theta} + \cos^2{\theta} =1.

Moreover, by the law of cosines, b^2=a^2+c^2-2a c \cos{\theta} which implies

\cos{\theta} = \frac{a^2+c^2-b^2}{2 a c},

and (1) becomes

A^2=\frac{1}{4}a^2 c^2 (1-\frac{(a^2+c^2-b^2)^2}{4 a^2 c^2})

=\frac{1}{4}a^2 c^2 \frac{4a^2 c^2-(a^2+c^2-b^2)^2}{4a^2 c^2}

=\frac{1}{16}(2a c +a^2+c^2-b^2)(2a c-a^2-c^2+b^2)





= s(s-a)(s-b)(s-c)