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How to Find the Area of a Triangle with 3 Coordinate Points

Updated on August 16, 2013
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TR Smith is a product designer and former teacher who uses math in her work every day.

calculate the area of a triangle with vertices given as coordinates.
calculate the area of a triangle with vertices given as coordinates.

In applied math and geometry, the bounaries of shapes are often given by spatial coordinates. If you have a triangle whose points are defined as (x, y)-coordinates, it is harder to apply the traditional formulas for triangular area such as Heron's formula, trig-based formulas, or the base-height formula. Instead, you can compute the area of the triangle subtractively by figuring the areas of simpler regions surrounding the triangle.

Step 1: Plot the Triangle and Draw a Rectangle Around It

Every triangle can be surrounded by a rectangle so that the points of the triangle lie on the edges of the rectangle. This creates several new shapes whose areas are easy to compute by hand. By way of example, we will start with a triangle whose Cartesian coordinates are (3, 3), (-2, 1), and (1, -4).

Draw a rectangle around the original triangle.
Draw a rectangle around the original triangle.

Step 2: Calculate the Area of the Rectangle

The area of a rectangle is simply the width times the length. In this example, the surrounding rectangle's area is 35 square units because it is 5 units wide and 7 units long.

Step 3: Calculate the Areas of the Surrounding Right Triangles

When you bound the original triangle by a rectangle, you produce several right triangles around the perimeter of the orignal triangle. The areas of these triangles are 1/2 times the product of the leg lengths (1/2 base times height). In this example we have three right triangles whose leg lengths are

This method always works, regardless of the coordinates of the triangle.
This method always works, regardless of the coordinates of the triangle.

7 and 2
5 and 3
2 and 5

This gives us areas of

0.5*7*2 = 7
0.5*5*3 = 7.5
0.5*2*5 = 5

The sum of these areas is 7 + 7.5 + 5 = 19.5 square units.

Step 4: Subtract the Areas of the Right Triangles from the Rectangle

When you subtract the combined area of the right triangles from the rectangle, the left over amount is the area of the original triangle. This is because the area of the original triangle plus the areas of the right triangles adds up to the total area of the rectangle.

In this example, the area of the rectangle is 35 square units and the total area of the right triangular regions is 19.5 square units. Therefore, the area of the original triangle is

35 - 19.5 = 15.5 square units.


Shortcut Formula

There is a shortcut formula that only requires the coordinates of the triangle. If the vertices are (A, B), (C, D), and (E, F), then the area of the triangle bound by these points is

0.5 * | AD+CF+EB-AF-CB-ED |

where | | is the absolute value function (i.e. positive values stay positive and negative values become negative). If we apply this formula to our example with coordinates (3, 3), (-2, 1), and (1, -4) we have

A = 3
B = 3
C = -2
D = 1
E = 1
F = -4

which gives

area = 0.5 * | 3 + 8 + 3 - (-12) - (-6) - 1|

= 0.5 * |31|

= 15.5

just as we obtained using the subtractive method.

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      Cass 3 years ago

      This is what I was looking for. I suppose you can do this with any shape with straight sides not just triangles. If you have a large collection of points (x1, y1), (x2, y2), ...., (xn, yn) can you also compute the area formulaically? That's what I'd really like to have so I could write a program.

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      TR Smith 3 years ago from Eastern Europe

      Yes, you can apply the rectangulation method to any polygon. As for a formula for more than 3 points, in theory there will be a formula so long as the points form a convex shape. But if some subset of points lies within the convex hull of another subset, there there won't be a single shape defined by the points. Instead there will be several different non-convex shapes with vertices at those points. In that case there won't be a well-defined formula for the area in terms of the points' coordinates.

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      Cass 3 years ago

      That makes sense, thanks.

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      andria 3 years ago

      if you have four points that define a convex shape and they are given in clockwise (or counterclockwise) order around the boundary then you can extend the 3-point forum to a 4-point formula by triangulating the region and applying the 3-point formula twice. Ex: if the 4 points are (a,b) (c,d) (e,f) (g,h) in circular order, then the area of the quadrilateral is

      ABS(ad+cf+eb-af-cb-ed)/2 + ABS(ah+gf+eb-af-gb-eh)/2

      (weird, form won't let me use absolute value bars)

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      Author

      TR Smith 3 years ago from Eastern Europe

      Thanks Andria!

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      Mike 3 years ago

      Is there a similar formula for the perimeter of a triangle given by three points? What about the centroid of the triangle? Thanks.

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      Author

      TR Smith 3 years ago from Eastern Europe

      You just need to apply the two-point distance formula to get the length of each side. Then add the three distances to get the perimeter. The distance formula for two points (a, b) and (c, d) is

      D = sqrt[(a-c)^2 + (b-d)^2]

      ----------------------------------------------

      If a triangle's vertices are (a, b), (c, d), and (e, f) then the x-coordinate of the centroid is

      x = (a+c+e)/3

      and the y-coordinate is

      y = (b+d+f)/3

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      loms 3 years ago

      works for me

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