Over at his blog, Dave Marain asked for the fourth coordinate, along with (a,b), (0,0) and (b,a) to give a parallelogram. And then he asked for the area (in at least five ways!)

Later, Dave clarified that he wanted (a+b,a+b). IOW, the origin was to be the second coordinate. And, WLOG assuming b > a > 0, the area of the parallelogram (actually a rhombus) is $b^2 - a^2$.

However, without specifying the order of the coordinates, we could have used (a-b, b-a) for the missing coordinate. Can we find the area of this parallelogram?

Can we find the area of any triangle from just its coordinates (0,0), (a,b), (c,d) ?  Any arbitrary quadrilateral? I think translating to get one point at the origin is ok, but rotating is not…

1. May 5, 2008 am31 4:53 am 4:53 am

Sounds like the shoelace algorithm.

May 5, 2008 am31 6:55 am 6:55 am

Uh, unless I’m missing something, isn’t this just the three step process, (i) “answer the triangle question”,
(ii) “w.l.o.g., assume one vertex of the polygon is at 0”, and
(iii) “do the completely obvious thing to use the first answer to solve the general problem”,
no?

May 5, 2008 pm31 3:05 pm 3:05 pm

There’s also Pick’s Theorem.

4. May 6, 2008 pm31 10:40 pm 10:40 pm

I hadn’t seen the shoelace algorithm before. It’s a nice little generalisation of the fact that the area of the parallelogram (0,0). (a,b). (c,d). (a+c, b+d) is determinant [a c \\ b d] (so triangles follow, and all other polygons follow that).

Pick’s Theorem is a truly splendid thing.

August 21, 2008 pm31 6:13 pm 6:13 pm

1. find for the area by green’s theorem or determinants
2. use pick’s theorem
3. to find for the vertex, […]

6. March 3, 2010 pm31 7:09 pm 7:09 pm

Im not sure what your question is, exactly, but if you want to find the area of any arbitrary simple polygon, given the coordinates of each vertex…

\$latex A = \frac{1}{2} | (x_1 y_2 – x_2 y_1 ) + (x_2 y_3 – x_3 y_2 ) + (x_3 y_4 – x_4 y_3 ) + \cdots (x_{n-1} y_n – x_n y_{n-1} ) + (x_n y_1 – x_1 y_n ) |

7. October 12, 2014 am31 6:40 am 6:40 am

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