Yeah, as we all know, finals, in most parts of the U.S are coming this week... On my geometry review, theres a few questions that I don't get, in particular being this one :

Determine the distance between the points (2, -2) and (5, -1).

a. square root of 10
b. 6
c. square root of 58
d. 2square root of 13.

anybody good at math?

HELP!
Its root 10.

Id draw a little picture for this,draw the two points and turn them into a triangle. Then use pythagoras
Ahh, this is simple. Make a triangle out of it.

It goes 3 across on the X axis, and goes up one on the Y axis. Your trying to work out the diagonal side which can be worked out with Pythagoras.

A^2+B^2=C^2

9+1=10

Square root ten.

EDIT: Every time I see American multi choice questions for Maths it makes me cringe...
Last edited by Craigo at Jun 1, 2008,
d = root of: 3^2 + 1^2

root 10.
distance formula
Distance Formula: X2-X1/Y2-Y1
Quote by ElMaco
My last pay check was £0 working 0 hours. I can't believe how easy it was
distance between 2 points formula
square root of ((x2-x1)^2 + (y2-y1)^2)
gives you square root of 9+1 or the square root of 10
lol im in 10th grade and ive knwn this for like 2 yrs now
First, draw it. Then use Pythagoras.

I wish we had multiple choice for maths
Quote by Minderbinder
Distance Formula: X2-X1/Y2-Y1

Erm, no.

...
sqrt(10)
Looking for my India/Django.
Quote by Minderbinder
Distance Formula: X2-X1/Y2-Y1

Isn't that the midpoint formula?

Quote by greatone_12
distance between 2 points formula
square root of ((x2-x1)^2 + (y2-y1)^2)
gives you square root of 9+1 or the square root of 10

This.
Quote by Minderbinder
Distance Formula: X2-X1/Y2-Y1

Secondly, it's Y2-Y1/X2-X1
Quote by rjdusa
Isn't that the midpoint formula?

This.

nah midpoint is
``````(X1+ X2), (Y1+Y2)
–––––––––––––––
2  ``````
Quote by Craigo
Mid point would be;

((X1+X2)/2),((Y1+Y2)/2)

Close enough. I havn't used that thing in years. Distance formula on the other hand... well TS try to remember that one. You'll probably need it.
Quote by rjdusa
Close enough. I havn't used that thing in years. Distance formula on the other hand... well TS try to remember that one. You'll probably need it.

Actually, there's no distance forumla. You apply basic geometrics into it. Pythagoras. About half the thread have said that...

Also, I worked out mid point for myself just with a little sense
Quote by Craigo
Actually, there's no distance forumla. You apply basic geometrics into it. Pythagoras. About half the thread have said that...

Also, I worked out mid point for myself just with a little sense

what do you call this then
square root of ((x2-x1)^2 + (y2-y1)^2)
actually it is pythagoras just modified to work with coordinate points
Quote by greatone_12
what do you call this then
square root of ((x2-x1)^2 + (y2-y1)^2)
actually it is pythagoras just modified to work with coordinate points

Also, you got the formula wrong ((X1-X2)^2+(Y1-Y2)^2)
Quote by Craigo

Also, you got the formula wrong ((X1-X2)^2+(Y1-Y2)^2)

the point is the magnitude not the sign and in the context of two points both are the same formula, x2-x1 just makes slightly more sense because it is standard to write differences as a second point minus the first, it makes no difference which you use though.
2-1=1
1-2=-1
(1)^2=(-1)^2
and I said it was a modified Pythagoras formula not that it was the original formula which would require a triangle drawn to make sense.
MULTIPLE CHOICE? ARE YOU ****ING KIDDING ME?

Goddammit, I want easy maths exams .
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Quote by greatone_12
the point is the magnitude not the sign and in the context of two points both are the same formula, x2-x1 just makes slightly more sense because it is standard to write differences as a second point minus the first, it makes no difference which you use though.
2-1=1
1-2=-1
(1)^2=(-1)^2
and I said it was a modified Pythagoras formula not that it was the original formula which would require a triangle drawn to make sense.

It's still Pythagoras theorem, no matter how you dress it up. All you've done is switched an expression with another expression.

And yes, you may be right, but X1-X2 and Y1-Y2 is conventionally correct
Quote by Minderbinder
Distance Formula: X2-X1/Y2-Y1

That is to find the tan of the angle, which equals the gradent (or pendient) of the line...
You find the coeficient of the first grade variable in the polinomic function with that (only if the polynomic function only has 1 grade, if not you have to get a curve and you have to find the gradent of various curves and find the centroid, and.....nevermind)
Quote by Craigo
It's still Pythagoras theorem, no matter how you dress it up. All you've done is switched an expression with another expression.

you can call it what you like but it works differently than Pythagoras even if thats where its derived from. its a line/segment formula not a triangle formula which would require the creation of new points and segments for the components(more appropriate in physics than in math), and since its a line/segment formula you can use it to solve straight lines(as in not diagonal) where creating a diagonal component would require you to rearrange Pythagoras, as far as i am concerned that makes it a different formula.
Last edited by greatone_12 at Jun 1, 2008,
Umm... i have a few more questions but I'm gonna scan them... do you guys mind helping me on a few more? thanks
I think its 80 because its a cyclic quadrilateral (basically meaning there are 4 point touching the edge of the circle) the opposite sides are therefore supplementary (they add to 180) just like in a triangle or line
These questions are very very easy - do a small amount of revision and then answer them yourself.
Quote by gonzaw
Lo, they give useless facts like two sides are equal, when the only thing you have to do is 180-Y....

Unless the problem continues...

Nah that doesnt work.

IDK, its been ages since I have done any geometry at all (Im into algebra and calculus and stuff) but by process of elim., I got 80. I got it because it looks about right

EDIT: On second thought, do what smb said!
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Last edited by zippidyduda at Jun 1, 2008,