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

[Petrus]

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

Spoiler

The area of a triangle is half of the area of parallelogram

 

[Ackbach]

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

Spoiler

$$[ \langle -4,1,0 \rangle- \langle 1,4,0 \rangle] \times [ \langle -4,1,0 \rangle- \langle -5,10,0 \rangle]= \langle -5,-3,0 \rangle \times \langle 1,-9,0 \rangle$$
$$= \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k}\\ -5 &-3 &0 \\ 1 &-9 &0\end{matrix} \right| = \hat{k}((-5)(-9)-(-3)(1))= \hat{k}(45+3)=48 \hat{k}.$$
Hence, the area of the triangle is $24$ units squared.

 

[MarkFL]

Re: Area

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

Spoiler

$$A=\frac{1}{2}|(-5+4)(4-1)-(1+4)(10-1)|=\frac{48}{2}=24$$

 

[Prove It]

Re: Area

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

Spoiler

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.

 

[TheEmptySet]

Re: Area

Spoiler

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