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I need to find the area of a parallelogram with vertices $(-1,-1), (4,1), (5,3), (10,5)$.

If I denote $A=(-1,-1)$, $B=(4,1)$, $C=(5,3)$, $D=(10,5)$, then I see that $\overrightarrow{AB}=(5,2)=\overrightarrow{CD}$. Similarly $\overrightarrow{AC}=\overrightarrow{BD}$. So I see that these points indeed form a parallelogram.

It is assignment from linear algebra class. I wasn't sure if I had to like use a matrix or something.

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  • $\begingroup$ Try drawing a picture. Work out the distance between each of the edges of the parallelogram (using Pythagoras' theorem). $\endgroup$ – Dan Rust Mar 16 '13 at 1:29
  • $\begingroup$ well cant I just use the area formula for a parallelogram and find the lengths of the edges? since it is linear algebra I wasnt sure if I had to like use a matrix or something. This just seemed like too simple of a question right? $\endgroup$ – D-Man Mar 16 '13 at 1:33
  • $\begingroup$ hint: draw a picture. The area is the magnitude of the cross product of certain two vectors. You will have to adjoin a third coordinate of $0$ to make the vectors three-dimensional $\endgroup$ – Stefan Smith Mar 16 '13 at 1:35
  • $\begingroup$ One definition of the area of a parallelogram is the determinant of the matrix with it's edge vectors as its columns. If you haven't been given/shown this, then you may need to prove it. $\endgroup$ – Dan Rust Mar 16 '13 at 1:35
  • $\begingroup$ Yeah, if you're not supposed to use cross products or you don't know what a cross product is, you'll have to do it another way which may look different but may actually be equivalent. $\endgroup$ – Stefan Smith Mar 16 '13 at 1:37
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Hint: the area of a parallelogram (see left-most image) is equal to the determinant of the $2\times 2$ matrix formed by the column vectors representing component vectors determined by the given points. $$A = \text{det}\,\left(\vec u \;\; \vec v\right)$$

The area of a parallelogram is also equal to the magnitutude of the cross product of the component vectors $\vec u, \vec v$ = $$|\vec u\times \vec v| = |\vec u|\,|\vec v|\sin \theta$$

where $\theta$ is the measure of the angle formed by the component vectors $\vec u, \vec v$.

Use your points to determine the component vectors $\vec u, \vec v$.

For parallelogram formed by $p_1, p_2, p_3, p_4$, put $\vec u = p_2 - p_1$, $\vec v = p_3 - p_1$ (where $p_4$ is the point opposite $p_1$, $p_2$ the point opposite $p_3$): $$\vec u = \langle 4 -(- 1), 1-(-1)\rangle = \langle 5, 2\rangle$$ $$\vec v = \langle 5 - (-1), 3 - (-1)\rangle = \langle 6, 4\rangle$$

So compute $$A = \det \begin{pmatrix} 5 & 6 \\ 2 & 4 \end{pmatrix},\;\;\text{or}\;\; A = \vert \vec u\times \vec v\vert$$


See the parallelogram whose area can be determined by component vectors $\vec u, \vec v\;\;$ (left of the image):

enter image description here

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  • $\begingroup$ "Hint: the area of a parallelogram (see left-most image) is equal to the determinant of the 2×2 matrix formed by the column vectors representing component vectors determined by the given points." But why? $\endgroup$ – Kenneth Worden Sep 8 '15 at 21:22
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Hints:

  1. The area of a parallelogram with side vectors $\bf a$ and $\bf b$ is $\det(\bf a\ \bf b)$.
  2. For a parallelogram $A,B,C,D$ its side vectors are e.g. $B-A$ and $C-A$.
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As the diagonal of a parallelogram $A(-1,-1),B(4,1),C(5,3),D(10,5)$ divides into two congruent triangles, which implies the triangles have same area.

So, the area of the parallelogram will be $2\cdot$ area of any one of $\triangle ABC,\triangle ADC, \triangle ABD, \triangle BCD$

For example, the area of $$\triangle ABC=\frac12\left|\det\begin{pmatrix} -1&-1&1 \\ 4&1&1 \\ 5&3&1\end{pmatrix}\right|$$

$$=\frac12\left|\det\begin{pmatrix} -1&-1&1 \\ 5&2&0 \\ 6&4&0\end{pmatrix}\right|\text{ applying} R_2'=R_2-R_1,R_3'=R_3-R_1$$

$$=\frac12\left|5\cdot4-6\cdot2\right|=4$$

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  • $\begingroup$ So because the area of triangle ABC was 4 and because it only represents half of the parallelogram the total area of this parallelogram would be 4 x 2 = 8? $\endgroup$ – D-Man Mar 16 '13 at 21:40
  • $\begingroup$ @D-Man, yes, you are right. $\endgroup$ – lab bhattacharjee Mar 17 '13 at 3:59

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