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.

2 thoughts on “Had Heron Known Analytic Geometry…

  1. Pingback: a x + b y + c = 0 : Why It Applies to All Straight Lines | Vroom

  2. Pingback: A Sprint to FTC | Vroom

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