Calculating the Area of a Parallelogram with Two Sets of Parallel Lines

[freeballa]

Homework Statement

This isn't really homework, but a not assigned problem out of my mathbook which is kinda confusing me..

#74. We have two pairs of parallel lines in R^2 defined by the linear equations below:

a1x + b1y = r1
a1x + b1y = s1
a2x + b2y = r2
a2x + b2y = s2

We assume that these lines enclose a parallelogram P. Find the very simplest formula for the area of P in terms of a1, b1, a2, b2, r1, r2, s1, s2.

Homework Equations

The Attempt at a Solution

Not sure how to attempt this problem because there are 4 equations for two lines?

 

[Dick]

There are FOUR lines. There are two pairs of parallel lines. The first two don't intersect and the last two don't intersect.

 

[freeballa]

Then the area P would be:

P = ||r1 x s1||
or..
P = ||r2 x s2||

Correct?

 

[Dick]

Then the area P would be:

P = ||r1 x s1||
or..
P = ||r2 x s2||

Correct?

r1 and s1 are numbers, not vectors. You can't take their cross product. You need to find some points on the intersection of the lines and take their difference to get vectors.

 

[freeballa]

I think I understand how to do it...

Find the intersection of the points:
a1x + b1y = r1 and a2x + b2y = r2
a1x + b1y = r1 and a2x + b2y = s2
a1x + b1y = s1 and a2x + b2y = r2
a1x + b1y = s1 and a2x + b2y = s2

Get the (x, y) for each of those points

Then subtract two sets by your selected new origin getting a and b.
Then put those into ||a x b||

But the question is asking me to put it into a formula; any help?

 

[Dick]

Look at this. http://en.wikipedia.org/wiki/Cramer's_rule

 

[freeballa]

Couldn't I just take the determinant of a1, b1, a2, b2?

Area = |a1b2 - a2b1|

 

[Dick]

Sure you can take the determinant, but what makes you think that has anything to do with AREA? It's obviously wrong. The a's and b's alone only tell you about the angles of the sides. The r's and s's determine how far apart they are.

 

[HallsofIvy]

It is true that if two intersecting sides of a parallelogram can be represented by $\vec{v}$ and $\vec{u}$ then the area of the rectangle is given by the length of $\vec{u}\times\vec{v}$.

It is also true that the area of a rectangular solid with sides intersecting at one corner given by vectors $\vec{u}$, $\vec{v}$, $\vec{w}$ is given by the "triple product", $|\vec{u}\cdot(\vec{v}\times \vec{w})|$ which is equal to the determinant formed with the vectors as columns or rows.

 

[freeballa]

a₁x + b₁y = r₁ and
a₂x + b₂y = r₂

X₁ = (r₁b₂-b₁r₂)/(a₁b₂-b₁a₂)
Y₁ = (a₁r₂-r₁a₂)/(a₁b₂-b₁a₂)

a₁x + b₁y = r₁ and
a₂x + b₂y = s₂

X₂ = (r₁b₂-b₁s₂)/(a₁b₂-b₁a₂)
Y₂ = (a₁s₂-r₁a₂)/(a₁b₂-b₁a₂)

a₁x + b₁y = s₁ and
a₂x + b₂y = r₂

X₃ = (s₁b₂-b₁r₂)/(a₁b₂-b₁a₂)
Y₃ = (a₁r₂-s₁a₂)/(a₁b₂-b₁a₂)

a₁x + b₁y = s₁ and
a₂x + b₂y = s₂

X₄ = (s₁b₂-b₁s₂)/(a₁b₂-b₁a₂)
Y₄ = (a₁s₂-s₁a₂)/(a₁b₂-b₁a₂)

Using a₁x + b₁y = r₁ and a₂x + b₂y = r₂ as midpoint:

A_X = X₂ - X₁
A_Y = Y₂ - Y₁

B_X = X₃ - X₁
B_Y = Y₃ - Y₁

Area = A_XB_Y - B_XA_Y

So, the simplest formula for P would be:

P = ((r₁b₂-b₁s₂)/(a₁b₂-b₁a₂) - (r₁b₂-b₁r₂)/(a₁b₂-b₁a₂))((a₁r₂-s₁a₂)/(a₁b₂-b₁a₂) - (a₁r₂-r₁a₂)/(a₁b₂-b₁a₂)) - ((s₁b₂-b₁r₂)/(a₁b₂-b₁a₂) - (r₁b₂-b₁r₂)/(a₁b₂-b₁a₂))((a₁s₂-r₁a₂)/(a₁b₂-b₁a₂) - (a₁r₂-r₁a₂)/(a₁b₂-b₁a₂))

 

[Dick]

If that is correct, then it's one answer. Have you checked it. I really doubt that it's the simplest. Since this isn't homework, I hope you aren't neglecting your real homework to pour time into this. Gotta confess, I actually don't know the answer.

 

[freeballa]

It hasn't been simplified, so it isn't in the simplest form. I am just checking if I did the correct steps as I have nothing to compare against to see if the answer is correct.

I am not neglecting my homework to do this.

 

[Dick]

Ok, it looks like you are doing the right sort of stuff. If you really want to check if it's correct, why don't you make some sample parallelograms with real numbers and check it out? I'm sorry to seem lazy here and not checking it myself. But, in fact, I am lazy. Sorry. The lord helps those who help themselves.

 

[freeballa]

Can anyone else confirm if I did this problem correctly?