Area of Triangle: Find Points A, B, & C

[adjacent]

Homework Statement

Find the area of triangle formed by the points ##A(5,2)## , ##B(4,7)## , ##C(7,-4)##

Homework Equations

Nah

The Attempt at a Solution

Is there any better way than finding the angle between lines and their lengths and then the area?

 

[SteamKing]

Homework Statement

Find the area of triangle formed by the points ##A(5,2)## , ##B(4,7)## , ##C(7,-4)##

Homework Equations

Nah

The Attempt at a Solution

Is there any better way than finding the angle between lines and their lengths and then the area?

That method sounds pretty tedious.

You can treat the triangle as a polygon with three sides. This article has a neat formula for finding the areas of polygons knowing only the coordinates of the vertices:

http://en.wikipedia.org/wiki/Polygon

Some words of caution:
1. the polygon must be traversed counterclockwise w.r.t. the origin to accumulate positive area.
2. the polygon must be closed, which implies that P1 = Pn, where Pk represents the coordinates of the kth vertex of the polygon.

 

[adjacent]

Is it this formula?

 

[BOAS]

Homework Statement

Find the area of triangle formed by the points ##A(5,2)## , ##B(4,7)## , ##C(7,-4)##

Homework Equations

Nah

The Attempt at a Solution

Is there any better way than finding the angle between lines and their lengths and then the area?

You could find the area via the vector cross product.

[itex]A = \frac{1}{2}|\vec{AB} \times \vec{AC}|[/itex]

 

[Fightfish]

Is it this formula?

Yup, it has a name, the "Shoelace algorithm".
http://en.wikipedia.org/wiki/Shoelace_formula

 

[adjacent]

You could find the area via the vector cross product.

[itex]A = \frac{1}{2}|\vec{AB} \times \vec{AC}|[/itex]

Oh. I haven't studied vector products yet. It's ##\frac{1}{2} AB \times AC \times \sin(\theta)## right?

 

[BOAS]

Oh. I haven't studied vector products yet. It's ##\frac{1}{2} AB \times AC \times \sin(\theta)## right?

##\frac{1}{2}|\vec{AB} \times \vec{AC}|## means to take the vector product of the two vectors, find the magnitude of the resultant vector (which gives the area of a parallelogram) and then multiply by 0.5 to get the area of a triangle.

In order to take the vector product you use a 'determinant' - Though you don't need to have studied matrices in order to perform this operation.

This page explains it better than I can;

http://mathworld.wolfram.com/CrossProduct.html

 

[SteamKing]

Is it this formula?

Yes, that's the one.

 

[SammyS]

Oh. I haven't studied vector products yet. It's ##\frac{1}{2} AB \times AC \times \sin(\theta)## right?

Yes, ##\ \frac{1}{2}|\vec{AB} \times \vec{AC}|=\frac{1}{2}\cdot |AB|\cdot |AC|\cdot \sin(\theta)\ .##

But doing the vector product using components, as suggested by BOAS eliminates the need to determine θ. In fact doing the vector product, or scalar product for that matter, via components allows you to determine θ.

 

[adjacent]

I will have to study advanced vectors then. When will it be covered?

 

[Ray Vickson]

Is it this formula?

Alternatively, you could use Heron's formula, which states that for a triangle with sides ##a, b, c## the area ##A## is given by
[tex] A = \sqrt{s(s-a)(s-b)(s-c)},\: \text{ where } \: s = \frac{a+b+c}{2} [/tex]
See, eg., http://en.wikipedia.org/wiki/Heron's_formula .

 

[Saitama]

Do you know about evaluating determinants? If so, there's a formula you might have seen.

Or maybe, does the following look familiar?
$$\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$$

 

[adjacent]

Do you know about evaluating determinants? If so, there's a formula you might have seen.

Or maybe, does the following look familiar?
$$\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$$

Is that some kind of determinant? Can you write that in the matrix form(If it has a matrix form)?

 

[Saitama]

Is that some kind of determinant?

I am not sure what you ask here. Did you comment on the formula I wrote?

The formula in the determinant form is:
$$\frac{1}{2}
\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{vmatrix}$$

 

[adjacent]

I am not sure what you ask here. Did you comment on the formula I wrote?

The formula in the determinant form is:
$$\frac{1}{2}
\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{vmatrix}$$

OMG, what is that? I have only studied to find the determinant of a 2x2 matrix
$$\begin{matrix}
a && b \\
c && d\\
\end{matrix}$$
Determinant is ad-bc

 

[Saitama]

OMG, what is that? I have only studied to find the determinant of a 2x2 matrix

Evaluating 3X3 determinants is very similar to 2X2 ones. You should look it up.

 

[adjacent]

Evaluating 3X3 determinants is very similar to 2X2 ones. You should look it up.

Why do we need a 3x3 matrix for calculating the area?

 

[verty]

If you know about direction cosines, you can just use the standard area formula without any of this fancy trickery.

Oh sorry, the question was, is there a better way? Not really, this is a very good way. But determinants are equally good, and a 2x2 can be used. There was a recent thread about this: here.

All I can say is, learn more linear algebra to use that method.

 

[verty]

I'll add, these are linear algebra topics, usually that is not precalculus math and therefore I recommend that people use the standard formula, or of course they can learn in advance to use something more advanced.