Freethought & Rationalism Archive

-RRH- · 04-01-2003, 09:45 PM

I'm trying to figure something out. See this diagram for my explanation.

I have the X and Y for both A and B. I want to find C and D.

Now, I remember doing something like this in a calculus course, but I haven't used it in a while, and I can't remember any key words that will find it through a search engine.

And thank you to anyone who can help me with this.

Jinto · 04-01-2003, 10:29 PM

Okay you have:

A = (Ax, Ay)
B = (Bx, By)

And are trying to find:

C = (Cx, Cy)
D = (Dx, Dy)

You know that the slope from the origin for points B and C are identical, so:

(Cy - 0)/(Cx - 0) = (By - 0)/(Bx - 0) --> Cy/Cx = By/Bx

and you know that the slope from C to A is perpendicular to this:

(Ay - Cy)/(Ax - Cx) = -Bx/By

So now you have two equations and two unknowns (Cx and Cy). Repeating the process for D, we know that the slope from D to A is parallel to the slope from the origin to B, so:

(Ay - Dy)/(Ax - Dx) = By/Bx

And since the slope from the origin to D is perpendicular:

Dy/Dx = -Bx/By

So now you have two equations and two unknowns for D as well. From here it's just another textbook algebra problem.

keyser_soze · 04-02-2003, 03:14 AM

Only at II!

Feather · 04-02-2003, 06:05 AM

PS: Jinto's post assumes the lines are orthogonal (i.e. that the quadrangle passing through A, C, D and the intersection of the black lines is a rectangle and that the black lines are orthogonal to one another).

-RRH- · 04-02-2003, 08:12 AM

Quote:

Originally posted by Feather
PS: Jinto's post assumes the lines are orthogonal (i.e. that the quadrangle passing through A, C, D and the intersection of the black lines is a rectangle and that the black lines are orthogonal to one another).

Right, I guess I really should have put in them little squares to show that is the case.

Thanks Jinto!

tensorproduct · 04-07-2003, 09:30 AM

Quote:

Originally posted by RRH

Now, I remember doing something like this in a calculus course, but I haven't used it in a while

A Calculus course!! as the solution given shows this can be done by simple school level geometry. Why would you think you need calculus?

Shadowy Man · 04-07-2003, 09:44 AM

It looks like using rotation matrices would be the easiest approach to this problem.

ImGod · 04-07-2003, 10:21 AM

Quote:

Originally posted by Shadowy Man
It looks like using rotation matrices would be the easiest approach to this problem.

I think the easiest method would be to draw a scale on each axis derived from one known point (and verified by the other known point). Then just read off the co-ordinates of the two unknown points.

Shake · 04-07-2003, 10:48 AM

Quote:

Originally posted by Shadowy Man
It looks like using rotation matrices would be the easiest approach to this problem.

Blech! Well, yeah, if you like that sort of thing! Not me. I can (well I could at one time) do calculus in 3D, but linear algebra ... YUCK! For some reason, I always had trouble with it. Maybe cause I know it shouldn't be that hard ... in fact, it gave me such an unpleasant experience, I doubt if I could go back to my old texts on it.

Jinto · 04-07-2003, 01:06 PM

Um, guys, I hate to point this out, but his problem has alrady been solved, so your discussion is moot.

04-01-2003, 09:45 PM	#1
-RRH- Veteran Member Join Date: Nov 2001 Location: Winnipeg Posts: 2,047	Math question I'm trying to figure something out. See this diagram for my explanation. I have the X and Y for both A and B. I want to find C and D. Now, I remember doing something like this in a calculus course, but I haven't used it in a while, and I can't remember any key words that will find it through a search engine. And thank you to anyone who can help me with this.

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04-01-2003, 10:29 PM	#2
Jinto Senior Member Join Date: Mar 2003 Location: SLC, UT Posts: 957	Okay you have: A = (Ax, Ay) B = (Bx, By) And are trying to find: C = (Cx, Cy) D = (Dx, Dy) You know that the slope from the origin for points B and C are identical, so: (Cy - 0)/(Cx - 0) = (By - 0)/(Bx - 0) --> Cy/Cx = By/Bx and you know that the slope from C to A is perpendicular to this: (Ay - Cy)/(Ax - Cx) = -Bx/By So now you have two equations and two unknowns (Cx and Cy). Repeating the process for D, we know that the slope from D to A is parallel to the slope from the origin to B, so: (Ay - Dy)/(Ax - Dx) = By/Bx And since the slope from the origin to D is perpendicular: Dy/Dx = -Bx/By So now you have two equations and two unknowns for D as well. From here it's just another textbook algebra problem.

04-02-2003, 03:14 AM	#3
keyser_soze Veteran Member Join Date: Mar 2003 Location: god's judge (pariah) Posts: 1,281	Only at II!

04-02-2003, 06:05 AM	#4
Feather Veteran Member Join Date: May 2002 Location: Gainesville, FL Posts: 1,827	PS: Jinto's post assumes the lines are orthogonal (i.e. that the quadrangle passing through A, C, D and the intersection of the black lines is a rectangle and that the black lines are orthogonal to one another).

04-07-2003, 09:44 AM	#7
Shadowy Man Veteran Member Join Date: Jun 2002 Location: A Shadowy Planet Posts: 7,585	It looks like using rotation matrices would be the easiest approach to this problem.

04-07-2003, 01:06 PM	#10
Jinto Senior Member Join Date: Mar 2003 Location: SLC, UT Posts: 957	Um, guys, I hate to point this out, but his problem has alrady been solved, so your discussion is moot.

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