Areas From Vectors

Tony computed the answers for some particular cases.

 

$$a d-b c\ \text{ (or } bc-ad\text{ if } bc>ad)$$

 

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Yingtong Yan solved the problem differently, using  stage 5 mathematics.

If we let $\mathbf{p}(x_1, y_1)$ and $\mathbf{q}(x_2, y_2)$ be our two vectors, then the cosinus of the angle between them is

$$\cos<\mathbf{p},\mathbf{q}>=\frac{\mathbf{p} \cdot \mathbf{q}}{|\mathbf{p}|\times |\mathbf{q}|}.$$

Therefore, $$\cos^2<\mathbf{p},\mathbf{q}>=\frac{(\mathbf{p} \cdot \mathbf{q})^2}{(|\mathbf{p}|\times |\mathbf{q}|)^2}$$

and, since $\sin^2<\mathbf{p},\mathbf{q}>+\cos^2<\mathbf{p},\mathbf{q}>=1$,

$$\sin^2<\mathbf{p},\mathbf{q}>=1 - \frac{(\mathbf{p} \cdot \mathbf{q})^2}{(|\mathbf{p}|\times |\mathbf{q}|)^2}= \frac{(|\mathbf{p}|\times |\mathbf{q}|)^2-(\mathbf{p} \cdot \mathbf{q})^2}{(|\mathbf{p}|\times |\mathbf{q}|)^2}.$$

Moreover, since the angle between $\mathbf{p}$ and $\mathbf{q}$ is smaller then $180^\circ$, then $\sin<\mathbf{p},\mathbf{q}>$ is greater then 0, and, hence, $\sin<\mathbf{p},\mathbf{q}> = \sqrt{\frac{(|\mathbf{p}|\times |\mathbf{q}|)^2-(\mathbf{p}\cdot\mathbf{q})^2}{(|\mathbf{p}|\times |\mathbf{q}|)^2} }$.

Using the fact that the parallelogram is made out of two congruent triangles of area $\frac{1}{2}\times\sin<\mathbf{p},\mathbf{q}>\times |\mathbf{p}|\times |\mathbf{q}|$, the area of the parallelogram is

$$\begin{aligned}

   A_{\text{parallelogram}} &= 2\times \frac{1}{2}\times\sin<\mathbf{p},\mathbf{q}>\times |\mathbf{p}|\times |\mathbf{q}|  \\

    &= \sqrt{\frac{(|\mathbf{p}|\times |\mathbf{q}|)^2-(\mathbf{p}\cdot\mathbf{q})^2}{(|\mathbf{p}|\times |\mathbf{q}|)^2} }\times |\mathbf{p|}\times |\mathbf{q}| \\

    &= \sqrt{ (|\mathbf{p}|\times |\mathbf{q}|)^2-(\mathbf{p}\cdot\mathbf{q})^2} \\

    &= \sqrt{(x_1^2+y_1^2)\times (x_2^2+y_2^2) - (x_1x_2+y_1y_2)^2} \\

   &=\sqrt{x_1^2y_2^2+x_2^2y_1^2-2x_1x_2y_1y_2} \\

    &=|x_1y_2-x_2y_1|

    \end{aligned}$$

Beth and Susie analyzed the special case when the parallelogram is a square. 

If the parallelogram is a square then $|\mathbf{p}|=|\mathbf{q}|$ and the area is $|\mathbf{p}|^2$. Since $|\mathbf{p}|=\sqrt{a^2+b^2}$, the area is $\sqrt{a^2+b^2}^2=a^2+b^2$.

Well done to everyone who sent in a solution. Keep exploring!