3

Three vertices are given. We can find the area using the determinant. Can someone explain it to me why the number will be a positive number, if vertices are chosen in anticlockwise direction.

Abhishek Bansal
  • 109
  • Is it ture? I was really wonder. – Bumblebee Sep 02 '14 at 12:43
  • Do you know about the cross product in $\mathbb{R}^3$? – Travis Willse Sep 02 '14 at 12:45
  • The thing is that I am teaching coordinate geometry, I have not started vectors. I have idea about cross product. cross product does give the area of a parallelogram. – Abhishek Bansal Sep 02 '14 at 12:46
  • Use the right hand rule and consider the determinant of the matrix given by $((B-A,0)^T, (C-A,0)^T, e_3^T)$ where $A,B,C$ are the vertices of the triangle going around anticlockwise and $e_3$ is the vector $(0,0,1)$ – Dan Rust Sep 02 '14 at 12:54

3 Answers3

1

The basis for the idea would be this picture (slightly modified from Wikipedia):

Slightly modified from the original wikipedia picture(https://en.wikipedia.org/wiki/Cross_product#mediaviewer/File:Cross_product_parallelogram.svg)

Here we would picture the given triangle as the one formed by $a,b$ and the heavy black line. The area of the parallelogram is twice the area of the triangle, so you would have to halve it at the end.

Suppose $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are the three points given counterclockwise from the base of $a$. Then the vector for $a$ is $[x_2-x_1,y_2-y_1]$, the vector for $b$ is $[x_3-x_1,y_3-y_1]$, and so the "determinant" picture of the cross product yields:

$$a\times b=\left|\begin{array}&i&j&k\\x_2-x_1&y_2-y_1&0\\x_3-x_1&y_3-y_1&0\end{array}\right|=\\ ((x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1))k$$

Thus the area of the triangle would be $\frac12((x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1))$

By keeping the points in counterclockwise order, it's guaranteed that the angle between $a$ and $b$ will be between $0$ and $\pi$, so the coefficient next to $k$ is positive. If the order is reversed, you'd just have to strip the negative sign off of the result.

rschwieb
  • 153,510
0

enter image description here $$\begin{align}\Delta&=ar(BFDA)+ar(BFEC)-ar(ACEF)\tag{1}\\\end{align}$$ Does that give you a hint? $$\Delta=\frac12\left|\begin{array}&x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{array}\right|$$

Since B is above AC, which is when ABC are clockwise. When B is below AC, which is when ABC are anticlockwise.

RE60K
  • 17,716
0

Using rschwieb's diagram, change the coordinate system so that $(x_1,y_1)$ is at the origin, and so that $x_2>0$ and $y_2 = 0$. (This obviously doesn't change the area). Then the three points are $(0,0)$, $(x_2,0)$, and $(x_3,y_3)$. Since the angle between $\mathbf{a}$ and $\mathbf{b}$ is strictly between $0$ and $\pi$, we know that $y_3 > 0$. Then $$\begin{vmatrix} x_2 & 0 \\ x_3 & y_3\end{vmatrix} = x_2y_3 > 0.$$

rogerl
  • 22,399