My Math Forum Analytic geometry help please

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 April 4th, 2010, 10:32 AM #1 Newbie   Joined: Mar 2010 Posts: 9 Thanks: 0 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)
 April 4th, 2010, 02:49 PM #2 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 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.
 April 4th, 2010, 03:20 PM #3 Senior Member     Joined: Feb 2010 Posts: 706 Thanks: 140 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
April 4th, 2010, 03:36 PM   #4
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 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?

 April 4th, 2010, 06:44 PM #5 Senior Member     Joined: Feb 2010 Posts: 706 Thanks: 140 Re: Analytic geometry help please If the vertices of the triangle are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ then the area of the triangle is given by Area = $\pm \frac{1}{2} \cdot \begin{vmatrix}1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \end{vmatrix}$ where the sign is determined by the direction in which the vertices are traversed.
 April 4th, 2010, 09:29 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 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.
April 5th, 2010, 01:07 AM   #7
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 Originally Posted by mrtwhs If the vertices of the triangle are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ then the area of the triangle is given by Area = $\pm \frac{1}{2} \cdot \begin{vmatrix}1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \end{vmatrix}$ 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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