matrix method to find area

[hassam]

wat's the matrix method to find area on co-ordinate axes... cn any 1 explain me that

 

[msk]

i don,t exactly remeber that....but lets see (1,2) (3,4) (2,6) [ = Modulus as in negative becomes positive
1/2 [ 1 3 2 1] [
[ 2 4 6 2]

now zig zag multiply these> 1/2[ (1x4x2x2) - (1x6x3x2)]
1/2[16-36)
1/2(20)
10 units.....Hope u get it

 

[Saturation]

I'm not sure if this is the matrix method, but I'm guessing this is it:

You have a shape on the graph. You need the coordinates of the Vertices.
Assume you have a triangle with vertices:

x1,y1
x2,y2
x3,y3

you write it down like this:

x1 x2 x3 x1 (Notice that you repeat the the 1st coordinate, in the end! This is important!)
y1 y2 y3 y1


The area then is:

(y1*x2) + (y2*x3) + (y3*x1) - (x1*y2) + (x2*y3) + (x3*y1)
DIVIDED by 2

This can be used for any shape. The difference is that you'll have more vertices.
MAKE SURE YOU REPEAT THE 1ST COORDINATE AT THE END!

 

[hassam]

cn u give me a link where a solved example is there..

 

[msk]

cn u give me a link where a solved example is there..

the way i told u is the right way...but still if u have the addmaths buk its given in that!

 

[Saturation]

Hey Saturation it's like this------>
0.5*[(y1*x2) + (y2*x3) + (y3*x1) - (x1*y2) - (x2*y3) - (x3*y1)]
And it only applies to a triangle with coordinate A(x1,y1) , B(x2,y2) & C(x3,y3)

I don't get your method, but if you want to find the area of a shape on a graph, the method I typed above works. And it works for other shapes aswell, not only triangles! (I'm not sure if there are limitations, but it definitely works for a rhombus, so I'm guessing other shapes too.) And btw I said I'm not sure whether this is the matrix method! Maybe yours is! :unknown:

I solved this Qs (No 8 (iii), using my method and got the answer:
http://www.xtremepapers.me/CIE/Internat ... _qp_13.pdf

Cheers! :Yahoo!: