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Geometric Proof of Area of Triangle Formula

Date: 04/30/2008 at 01:01:17
From: Gerry
Subject: Proof of Area of Triangle Formula

I'm trying to prove the formula that the area of a triangle with co-
ordinates (0,0),(x1,y1) and (x2,y2) is 1/2(x1y2 - x2y1) without using
determinants.  Am sure I recall an elegant way to do this from when I
was in school but that was 20 years ago so it escapes me now.

I seem to end up with a jumble of algebra when I try to "brute force" 
the proof but I'm sure there is a cleaner way.

If one vertex is (0,0), another (x1,y1) and another (x2,y2) then 
trying to use the 1/2 * base * height formula gets me the slope of 
one side = y1/x1 so slope of perpendicular is -x1/y1.  Can use this 
slope and the point (x2,y2) to get equation of perpendicular height
line and then find point of intersection with line y = y1/x1 x by
solving simultaneously.  Could then use this point and (c2,y2) in the 
formula for the distance between two points sqrt((x2-x1)^2 + (y2-y1)
^2)) on top of this but the algebra gets pretty horrendous pretty 

Any thoughts appreciated.

Date: 04/30/2008 at 15:51:37
From: Doctor Peterson
Subject: Re: Proof of Area of Triangle Formula

Hi, Gerry.

It can be done using the vector cross-product; but maybe that's too 
close to determinants for your taste.

Here's a nice geometrical way, though it might take several cases to 
make it complete. Take this triangle:

  |      B(x2,y2)
  |     /  \
  |    /     \
  |   /        \
  |  /           A(x1,y1)
  | /       /
  |/   /

We can add in the vertical lines to A and B:

  |      B(x2,y2)
  |     /: \
  |    / :   \
  |   /  :     \
  |  /   :       A(x1,y1)
  | /    :  /    :
  |/   / :       :

Now go around the triangle from O to A to B and back to O, looking 
at the "shadows" of each edge, namely triangle OAD, trapezoid ABCD, 
and triangle BOC.  If we take the first of these as negative (since 
it is below triangle OAB), and the others as positive (since they 
include parts of OAB), and add the areas, what do we get?

  -Area(OAD) + Area(ABCD) + Area(BOC) = Area(OAB)

That is, we are subtracting the area of triangle OAD from 
quadrilateral ODAB which contains it, leaving only the desired 
triangle.  The details will vary according to the relationships of 
the points, but it all works out for every case.  There's probably a 
nicer (but perhaps hard to express) way to cover every case at once, 
but this satisfies me informally.

Now express these in terms of the coordinates:

  Area(OAB) = -[x1 y1]/2 + [x1 - x2][y1 + y2]/2 + [x2 y2]/2

            = [-x1 y1 + x1 y1 + x1 y2 - x2 y1 - x2 y2 + x2 y2]/2
               \____________/                 \_____________/

            = [x1 y2 - x2 y1]/2

This will be positive if OAB is traversed counterclockwise, and 
negative if it is traversed clockwise.  To be sure to get the 
(positive) area, take the absolute value.  And that's your formula.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum 

Date: 04/30/2008 at 19:15:59
From: Gerry
Subject: Thank you (Proof of Area of Triangle Formula)

Thanks very much!  That was exactly the proof I was looking for.

Associated Topics:
High School Triangles and Other Polygons

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