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April 4th, 2010, 10:32 AM   #1
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Analytic geometry help please

Help please,

Calculate the area of the triangle with corners A = (3, -2), B = (2, 1) and C = (-1, 3)

I know what to do but I'm not quite sure how to; I want to caculate the lengths between ab, bc using the formula square root of (y2-y1)^2+(x2-x1)^2.

So:
ab: sqroot ((1+2)^2+(2-3)^2) = sqroot (9+1) = sqroot (10)
bc: sqroot ((3-1)^2+(-1-2)^2) = sqroot (4+9) = sqroot (13)
ac: sqroot ((3+2)^2+(-1-3)^2) = sqroot (25+16) = sqroot (41)
(I'm not sure if these are correct)
since it's neither a right triangle nor are the feet the same length I can't just use Pythagoras or take base/2 and then pythagoras... so I don't really know what to do now.. I don't even know if I started out right on this because it seems a tad complicated to be the third task in the first practise-test of the course in the book (the first ones were a breeze)
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April 4th, 2010, 02:49 PM   #2
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Re: Analytic geometry help please

There are several ways to proceed, and I can't begin to guess which ones you would know. The easiest way, which I can only hope is what you've learnt, would be to use Heron's formula.

Or, you could note that if you split a triangle into two right triangles, you can come up with a formula (which happens to be Heron's formula), as well.

I don't really know if that's helpful or not.
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April 4th, 2010, 03:20 PM   #3
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Re: Analytic geometry help please

If this is a question from a standard analysis or pre-calculus course then you might want to look in your notes or your book for a determinant form. This will allow you to find the area from the coordinates of the vertices by just evaluating a particular 3 x 3 determinant.

MrTwhs
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April 4th, 2010, 03:36 PM   #4
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Re: Analytic geometry help please

Quote:
Originally Posted by mrtwhs
If this is a question from a standard analysis or pre-calculus course then you might want to look in your notes or your book for a determinant form. This will allow you to find the area from the coordinates of the vertices by just evaluating a particular 3 x 3 determinant.

MrTwhs
This strikes me as unlikely, as the determinant of a 3x3 matrix represents a three dimensional volume, and would be 0 for any triangle. Perhaps I am remiss?
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April 4th, 2010, 06:44 PM   #5
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Re: Analytic geometry help please

If the vertices of the triangle are then the area of the triangle is given by


Area =

where the sign is determined by the direction in which the vertices are traversed.
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April 4th, 2010, 09:29 PM   #6
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Let O = (0, 0) and D = (7, 0).

OC, AB are parallel, so triangles ABC, ABO have the same area.
OB, AD are parallel, so triangles ABO, DBO have the same area.

In triangle DBO, choose OD (which has length 7) as base, so that the corresponding height is 1.
Area triangle ABC = area triangle DBO = base height / 2 = 7 1 / 2 = 3.5.
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April 5th, 2010, 01:07 AM   #7
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Re: Analytic geometry help please

Quote:
Originally Posted by mrtwhs
If the vertices of the triangle are then the area of the triangle is given by


Area =

where the sign is determined by the direction in which the vertices are traversed.
What an interesting consideration. I had not seen this before, but a quick check showed this to be true. And, I imagine, very useful. I stand corrected, and thank you.
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