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 March 7th, 2012, 03:31 AM #1 Newbie   Joined: Mar 2012 Posts: 4 Thanks: 0 I want to find the distance in algebra find it Find the distance between the points (-4 , -5) and (-1 , -1)in algebra please give answer
 March 7th, 2012, 03:51 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,502 Thanks: 1739 5
 March 7th, 2012, 06:13 AM #3 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: I want to find the distance in algebra find it I keep getting 5.07 for some reason. I'd better go edit the wikipedia page for pythagorean triples!
 March 7th, 2012, 06:15 AM #4 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: I want to find the distance in algebra find it It appears that our friend kevin is no longer with us :P
March 7th, 2012, 01:05 PM   #5
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Re: I want to find the distance in algebra find it

Quote:
 Originally Posted by The Chaz I keep getting 5.07 for some reason. I'd better go edit the wikipedia page for pythagorean triples!
Strange answer - the right triangle involved is (3,4,5).

March 7th, 2012, 01:18 PM   #6
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Quote:
 Originally Posted by The Chaz I keep getting 5.07 for some reason.
You should get exactly 5 once you've made the final payment for the calculator. In the meanwhile, kevinman can return from the dead if no longer linking to a tutoring site in every post.

 March 7th, 2012, 08:09 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 510 Math Focus: Calculus/ODEs Re: I want to find the distance in algebra find it Based on the trivial nature of the problems posted, I doubt this user will want to come back if not allowed to spam our forum. The easiest, most straightforward way by far to find the given distance above is to employ a translation, then a rotation of axes. Translation: X = x + 4 Y = y + 5 Now the points are: (0,0), (3,4) See how we craftily made the point to the lower left the new origin? Rotation where ? = arctan(4/3) to put both points on the X-axis: X = x·cos(?) + y·sin(?) = (3/5)x + (4/5)y Y = -x·sin(?) + y·cos(?) = -(4/5)x + (3/5)y Now we have reduced the problem to one dimension and the points are (0), (5) Now the distance is d = |5 - 0| = 5. Trying to find the distance between two arbitrary points in two-dimensional space is far too messy to contemplate.
 March 8th, 2012, 05:15 AM #8 Global Moderator   Joined: Dec 2006 Posts: 19,502 Thanks: 1739 You might like Mathematics Made Difficult by Carl E. Linderholm . . . random quotes: "a little topological group representation theory is not amiss if you happen to end up a quantum mechanic, repairing other people's quanta when they begin to wear out" and "in a mathematical discussion, a chicken is often as much use as a man and makes comments of about equal intelligence." It contains plenty of mathematics, but has the occasional mistake, such as "The square root of 7$\small\frac{74}{2601}$ is obviously 2$\small\frac{13}{19}$."
March 8th, 2012, 05:53 AM   #9
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Re: I want to find the distance in algebra find it

Quote:
Originally Posted by mathman
Quote:
 Originally Posted by The Chaz I keep getting 5.07 for some reason. I'd better go edit the wikipedia page for pythagorean triples!
Strange answer - the right triangle involved is (3,4,5).
Guys.
Seriously.
Give me some credit, here!

(Or is it wrong of me to slip wrong answers in when dealing with a spammer?)

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