Finding area of triangle defined by two vectors

Question

I am trying to prove that the area of a triangle defined by two vectors $\left(x_1, x_2\right)$ , $(y_1, y_2)$ is $ \frac{|x_1y_2 - x_2y_1|}{2}$. The only way I've been able to do it so far is by cases. For example, if all of the components are positive, then the result follows from a simple geometric argument. I used a similar argument for the case where one of the components is negative, and so on. Is there a better way to do this that doesn't involve anything beyond high school level math?

Answer 1

Let $O(0,0)$, $A(x_1,x_2)$ and $B(y_1,y_2)$ be vertexes of our triangle.

Now, if $x_1=y_1$ we get that the area is $$\frac{|x_1|\cdot|x_2-y_2|}{2}=\frac{|x_1y_2-x_2y_1|}{2}.$$ If $x_1\neq y_1$ we can use the following formula.

Let $ax+by+c=0$ is an equation of line $l$. Then the distance between $A(x_1,y_1)$ and $l$ is $$\rho(A,l)=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$$

The equation of the line $AB$ is $$y-y_2=\frac{x_2-y_2}{x_1-y_1}(x-y_1)$$ or $$(x_2-y_2)x+(y_1-x_1)y+x_1y_2-x_2y_1=0.$$ The altitude of $\Delta OAB$ to $AB$ is $$\rho(O,AB)=\frac{|(x_2-y_2)\cdot0+(y_1-x_1)\cdot0+x_1y_2-x_2y_1|}{\sqrt{(x_2-y_2)^2+(y_1-x_1)^2}}=$$ $$=\frac{|x_1y_2-x_2y_1|}{AB},$$ which says that the area is $$\frac{\frac{|x_1y_2-x_2y_1|}{AB}\cdot AB}{2}=\frac{|x_1y_2-x_2y_1|}{2}$$ and we are done!

Answer 2

we have $$\vec{a}\times \vec{b}=[a_1,a_2,a_3]\times[b_1,b_2,b_3]=[a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1]$$ $$a_3=0$$ and $$b_3=0$$ in your case

Answer 3

Alternatively you can show that: $$ \begin{align} ||X \times Y||^2 =& ||X||^2\,||Y||^2 - |X \cdot Y|^2 \\ &\text{that resembles high school trigonometry} \\ =& ||X||^2\,||Y||^2 - ||X||^2 \, ||Y||^2 \cdot \cos^2 \theta \end{align} $$ where $X = (x_1, x_2)$ and $Y = (y_1, y_2)$

By doing some multiplications you get: $$||X||^2\,||Y||^2 - |X \cdot Y|^2 = (x_1^2+y_1^2)\cdot(x_2^2+y_2^2) - (x_1x_2 + y_1y_2)^2 = x_1^2y_2^2+y_1^2x_2^2 - 2x_1x_2y_1y_2$$ On the other hand: $$||X \times Y||^2 = (x_1y_2-x_2y_1)^2 = x_1^2y_2^2 +x_2^2y_1^2 - 2x_1x_2y_1y_2 $$ which concludes the demonstration.

The first equality shows that $||X \times Y||$ is area of a rectangle formed by vectors $X$ and $Y$. Basically what you are doing here is calculating $||X||\,||Y|| \cdot \sin\theta = b\cdot h$ of a rectangle (draw it).

The solution to your problem follows: $$A_{triangle} = \frac{A_{rectangle}}{2} = \frac{||X \times Y||}{2} = \frac{|x_1y_2-x_2y_1|}{2}$$