How do you find the area of a triangle from the co-ordinates of its vertices, without having to calculate the length of the triangles sides?

2 Answers
Jun 12, 2018

Answer:

See below.

Explanation:

Finding the area of a triangle from its co-ordinates.

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Using diagram.

We are looking for area of #Delta ABC# with co-ordinates:

#A=(x_1,y_1), B= (x_2 ,y_2), C=(x_3 , y_3)#

We form perpendiculars BD, AD and CF to the x axis.

We then notice that:

# "Area of" DeltaABC= "area of trapezium " DBAE + "area of trapezium " EACF - "area of trapezium " DBCF#

#:.#

#"Area " Delta ABC=1/2(y_2+y_1)(x_1-x_2) + 1/2(y_1+y_3)(x_3-x_1)-1/2(y_2+y_3)(x_3-x_2)#

Factor out #1/2# and expand:

#1/2[(y_2+y_1)(x_1-x_2) + (y_1+y_3)(x_3-x_1)-(y_2+y_3)(x_3-x_2)]#

#1/2[-y_1x_2+y_2x_1+y_1x_3-y_3x_1-y_2x_3+y_3x_2]#

Collect terms containing #x_1#

#y_2x_1-y_3x_1=>x_1(y_2-y_3)#

Terms containing #x_2#

#y_3x_2-y_1x_2=>x_2(y_3-y_1)#

Terms containing #x_3#

#y_1x_3-y_2x_3=>x_3(y_1-y_2)#

We now have a formula:

#1/2|[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]|#

The absolute bars ensure that the area will be positive, whatever order we take them in.

This is a much better method than calculating the length of the sides, which invariably will be in surd form, making further calculations difficult. There isn't a lot of information on this and similar methods, and the reason for answering my own question was to show this method. I have to give credit to contributors Dean R and Binyaka C for pointing me in the direction of this and similar methods.

Jun 17, 2018

Answer:

A triangle with vertices #(x_1, y_1), (x_2, y_2)# and #(x_3,y_3)# has area #S# given by the Shoelace formula,

#S = 1/2 | x_1 y_2 - x_2 y_1 + x_2 y_3 - x_3 y_2 + x_3 y_1 - x_1 y_3 |#

We can also write it as

#S = 1/2 | (x_1 - x_3)(y_2 - y_3) - (x_2 - x_3)(y_1-y_3) |#

Explanation:

Let's imagine a triangle with vertices #(0,0), (a,b), (c,d).# For now think of #(a,b)# and #(c,d)# in the first quadrant, #0< c < a#, #0 < b < d.#

The triangle's area #S# is the #a \times d # rectangle less the three right triangles.

# S = ad - 1/2 ab - 1/2 cd - 1/2 (d-b)(a-c)#

# S = ad - 1/2(ab + cd + ad -cd - ab + bc) #

# S = 1/2 (ad - bc)#

The area is half the cross product. This is the signed area; if we had taken the points in the other order we'd get negative the area.

#S = 1/2 | ad - bc | #

For an arbitrary triangle we translate #(x_3, y_3)# to the origin,

We substitute #a = x_1 - x_3, b=y_1-y_3,## c=x_2-x_3, d=y_2-y_3# giving

#S = 1/2 | (x_1 - x_3)(y_2 - y_3) - (x_2-x_3)(y_1-y_3) | #

That's one of our formulas. Multiplying it out, we get the standard form of the Shoelace Formula,

#S = 1/2 | x_1 y_2 - x_2 y_1 + x_2 y_3 - x_3 y_2 + x_3 y_1 - x_1 y_3 |#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

When we have three or more dimensions for the vertices we should do it this way:

Given vertices #(x_1, x_2, ..., x_n),# #(y_1, y_2, ..., y_n), (z_1, z_2, ..., z_n)# calculate

# A = (x_1 - y_1)^2 + (x_2 - y_2)^2 + ... + (x_n - y_n)^2#

# B = (y_1 - z_1)^2 + (y_2 - z_2)^2 + ... + (y_n - z_n)^2#

# C = (z_1 - x_1)^2 + (z_2 - x_2)^2 + ... + (z_n - x_n)^2#

# 16S^2 = 4AB - (C-A-B)^2#

but I won't derive that one now.