Monthly Archives: June 2017

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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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

|\frac{1}{2}D|\quad\quad\quad\quad\quad\quad\quad\quad\quad(1)

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.

Therefore,

A^2=s(s-a)(s-b)(s-c).

where

a^2=(x_2-x_3)^2+(y_2-y_3)^2,

b^2=(x_1-x_3)^2+(y_1-y_3)^2,

c^2=(x_1-x_2)^2+(y_1-y_2)^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

A^2-B^2=0.

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=0,

i.e.,

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:

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

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})

i.e.,

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)

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

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

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

=\frac{1}{16}\;2^4\;(\frac{a+b+c}{2})(\frac{a+b+c}{2}-a)(\frac{a+b+c}{2}-b)(\frac{a+b+c}{2}-c)

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

Hence,

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