I have this simple (high school level) exercise:
The $ABCD$ quadrilateral is a parallelogram of vertices $A(0;0)$, $B(20;10)$ and $D(10,y)$. If your area is $600$, what is the measure of $y$?
I have thought that the area of the parallelogram is:
$$\mathcal{A}_{\text{area}}=|ad-bc| \tag{1}$$
Hence I will have as solutions:
EDIT by comment:
$$600=|ad-bc|=|100-20(y+10)|\iff y=-35\quad \mathrm{or}\quad \:y=25.$$
but the solution is $y=35$ and this way not give me the solution of the problem, why?
Question:
Why with the $(1)$ have I not to find the solution?
Is there a simpler proof of $(1)$ like to that have I put?
Proof: If I have two vectors $\bar a$ and $\bar b$ then area of a parallelogram is:
$$|\bar a\times \bar b|=ab|\sin(\theta)|=ab\sqrt{1-\cos^2(\theta)}\tag{2}$$
Now $$\cos(\theta)=\frac{\bar a\cdot\bar b}{ab} \tag{3}.$$
Replacing the $(3)$ in the $(2)$ we get:
$$|\bar a\times \bar b|=ab\sqrt{1-\cos^2(\theta)}=ab\sqrt{1-\left(\frac{\bar a\cdot\bar b}{ab}\right)^2}$$
$$=\sqrt{a^2b^2\left(1-\left(\frac{\bar a\cdot\bar b}{ab}\right)^2\right)}=\sqrt{(\bar a \cdot \bar a)(\bar b \cdot \bar b)-(\bar a\cdot\bar b)^2}. \tag{4}$$
If $\bar a=(a_1,a_2)$ and $\bar b=(b_1,b_2)$ we have from $(4)$ that:
$$=\sqrt{(a_1^2+a_2^2)(b_1^2+b_2^2)-(a_1b_1+a_2b_2)^2}$$
or $$=\sqrt{(a_1b_2-a_2b_1)^2}=|a_1b_2-a_2b_1|$$
This gives $(1)$. $\square$
Best Answer
Let $A(0,0)$, $B(x_1,y_2)$, $C(x_2,y_2)$ be three points in a standard Cartesian plane. We want to compute the area of the triangle $ABC$.
The line $BC$ has equation $$ (y_1-y_2)(x-x_2)-(x_1-x_2)(y-y_2)=0 $$ so the distance of this line from the origin is $$ \frac{|(y_1-y_2)(0-x_2)-(x_1-x_2)(0-y_2)|}{\sqrt{(y_1-y_2)^2+(x_1-x_2)^2}}= \frac{\lvert-x_2y_1+x_2y_2+x_1y_2-x_2y_2\rvert}{\sqrt{(y_1-y_2)^2+(x_1-x_2)^2}}= \frac{\lvert x_1y_2-x_2y_1\rvert}{\sqrt{(y_1-y_2)^2+(x_1-x_2)^2}} $$ Then the area of the triangle is obtained as half of this distance times the lenght of $BC$, which is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$, so finally the area is $$ \frac{1}{2}\lvert x_1y_2-x_2y_1\rvert= \frac{1}{2}\left|\det\begin{bmatrix} x_1 & x_2 \\ y_1 & y_2 \end{bmatrix}\right| $$ If we add a fourth vertex making a parallelogram, just double this area.
If instead the point $A$ has coordinates $A(x_0,y_0)$, a translation is sufficient and the area is $$ \frac{1}{2}\left|\det\begin{bmatrix} x_1-x_0 & x_2-x_0 \\ y_1-y_0 & y_2-y_0 \end{bmatrix}\right| $$ This also admits a representation as a $3\times3$ determinant, because $$ \det\begin{bmatrix} x_0 & x_1 & x_2 \\ y_0 & y_1 & y_2 \\ 1 & 1 & 1 \end{bmatrix} = \det\begin{bmatrix} 0 & x_1-x_0 & x_2-x_0 \\ 0 & y_1-y_0 & y_2-y_0 \\ 1 & 1 & 1 \end{bmatrix} =\det\begin{bmatrix} x_1-x_0 & x_2-x_0 \\ y_1-y_0 & y_2-y_0 \end{bmatrix} $$