How Can You Calculate the Area of a Triangle in an Orthogonal Coordinate System?

In summary, the conversation discusses different methods for finding the area of a triangle with vertices in an orthogonal coordinate system, including using vector cross product and Heron's Formula. The formula developed in a math help forum is also mentioned as a possible solution.
  • #1
Petrus
702
0
In an orthogonal cordinate system determine the area of the triangle with vertices in \(\displaystyle (-4,1)\), \(\displaystyle (1,4)\) and \(\displaystyle (-5,10)\) There is prob many way to solve it so go ahead with your method:) I made a hint for vector method.
Hint
The area of a triangle is half of the area of parallelogram
 
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  • #2
Re: Area

The vector cross product, I believe (or the magnitude thereof) gives the area of the corresponding parallelogram. Hence, divide by two to get the answer. I get
$$[ \langle -4,1,0 \rangle- \langle 1,4,0 \rangle] \times [ \langle -4,1,0 \rangle- \langle -5,10,0 \rangle]= \langle -5,-3,0 \rangle \times \langle 1,-9,0 \rangle$$
$$= \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k}\\ -5 &-3 &0 \\ 1 &-9 &0\end{matrix} \right| = \hat{k}((-5)(-9)-(-3)(1))= \hat{k}(45+3)=48 \hat{k}.$$
Hence, the area of the triangle is $24$ units squared.
 
  • #3
Re: Area

Using the formula I developed http://www.mathhelpboards.com/f49/finding-area-triangle-formed-3-points-plane-2954/, we find:

\(\displaystyle A=\frac{1}{2}|(-5+4)(4-1)-(1+4)(10-1)|=\frac{48}{2}=24\)
 
  • #4
Re: Area

Petrus said:
In an orthogonal cordinate system determine the area of the triangle with vertices in \(\displaystyle (-4,1)\), \(\displaystyle (1,4)\) and \(\displaystyle (-5,10)\) There is prob many way to solve it so go ahead with your method:) I made a hint for vector method.
Hint
The area of a triangle is half of the area of parallelogram

If you get really desperate, you could work out the length of each side of the triangle by evaluating the distance between each set of points, and then use Heron's Formula.
 
  • #5
Re: Area

The good old fashion method. The slope between the points (-4,1) and (1,4) is \(\displaystyle \frac{3}{5}. \)So the slope of a line perpendicular to it is \(\displaystyle -\frac{5}{3}\)This gives the two equations \(\displaystyle y-1=\frac{3}{5}(x+4)\) the equation of the line perpendicular is \(\displaystyle y-10=-\frac{5}{3}(x+5)\) Solving the system for the point of intersection gives \(\displaystyle \left(-\frac{13}{17},\frac{50}{17} \right)\) The length of each line segment can now be calculated \(\displaystyle b=\int_{-4}^{1}\sqrt{1+(\frac{3}{5})^2}dx=\frac{24}{17} \sqrt{34}\\ h=\int_{-5}^{-\frac{13}{17}}\sqrt{1+(-\frac{5}{3})^2}dx=\sqrt{34}\\\) This gives the area of the triangle \(\displaystyle A=\frac{1}{2}bh=24\)
 

FAQ: How Can You Calculate the Area of a Triangle in an Orthogonal Coordinate System?

What is the formula for finding the area of a triangle?

The formula for finding the area of a triangle is: A = 1/2 * base * height, where A is the area, base is the length of the triangle's base, and height is the length of the triangle's height.

What is the difference between the base and height of a triangle?

The base of a triangle is the longest side of the triangle, while the height is the perpendicular distance from the base to the opposite vertex.

Can the area of a triangle be negative?

No, the area of a triangle cannot be negative. It is always a positive value.

Does the shape of a triangle affect its area?

Yes, the shape of a triangle can affect its area. For example, two triangles with the same base and height will have the same area, but two triangles with the same base and different heights will have different areas.

How can I use the area of a triangle in real life?

The area of a triangle can be used in many real-life situations, such as determining the amount of paint needed to cover a triangular wall, calculating the size of a triangular garden bed, or finding the area of a roof with a triangular shape. It is also used in various fields of science, including architecture, engineering, and physics.

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