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I would like to coin together certain properties of

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I would like to coin together certain properties of [#permalink] New post 28 Dec 2006, 20:27
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I would like to coin together certain properties of Coordinate Geometry (from the basics) that I learnt from various sources, some even from the forum. Would appreciate your help in adding more or correcting these.

SLOPE:

slope m=(y1-y2)/(x1-x2)

*A straight line with a -ve slope passes through II and IV quadrants
A straight line with a +ve slope passes through I and III quadrants

*If the slope is 1 the angle formed by the line is 45 degrees

*If the slope of a line is n, the slope of a line perpendicular to it is its -ve reciprocal, -1/n

*If a line is horizontal, slope=0, equation is y=b

*If a line is vertical, slope is not defined, equation is x=a, where a is x-intercept.

*Parallel lines have same slope.

EQUATION:

Equation of a circle is (x - a)^2 + (y-b)^2 = r^2, where (a,b) is the center and r is the radius

Equation of a circle is x^2+y^2=r^2 if (0,0) is the center

Equation of a line is y=mx+b, where m=slope and b=y intercept

* y intercept is the value of y when x is 0
x intercept is the value of x when y is 0

* Points that solve the equation of a line are in the same line

*Given a point and slope, equation of the line can be found

*Given the equation, x and y intercepts can be found


DISTANCE between two points = sqrt[(x1-x2)^2+(y1-y2)^2]

MIDPOINT: xm=(x1+x2)/2 ym=(y1+y2)/2

I know, the properties do not end here. Hope you guys join in.

Last edited by Sumithra on 16 Jan 2007, 12:46, edited 3 times in total.
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 [#permalink] New post 29 Dec 2006, 06:35
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Thank you tennis ball. Edited my previous post.

Can somebody explain 'functions'?
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 [#permalink] New post 16 Jan 2007, 12:53
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Would appreciate if someone could shed some light on how to view an equation/inequality in xy-plane, for both straight lines and curves?
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 [#permalink] New post 16 Jan 2007, 15:16
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Sumithra wrote:
Would appreciate if someone could shed some light on how to view an equation/inequality in xy-plane, for both straight lines and curves?


For lines, curves or any functions represented on an XY Plan, the concepts remain similar.

o y > f(x)
On the XY plan, the matching values of y are "above" the draw of y = f(x).

To know where is the concerned region, u can use 1 value of x : x=0 (if definied on 0), calculate f(0) and then look where is a value of y such that y > f(0). U will be "above" the f(x).

Ex 1 : For the line f(x) = 3*x+1, y > f(x) is in green (fig 1). f(0) = 1 so u know that the point (0,2) is in good region. Now, we can draw the region limited by the line and containing (0,2).

Ex 2 : For the line f(x) = -2*x-1, y > f(x) is in green (fig 2). f(0) = -1 so u know that the point (0,0) is in good region.

Ex 3 : For the line f(x) = x^2 -2*x -1, y > f(x) is in green (fig 3). f(0) = -1 so u know that the point (0,0) is in good region.



o y < f(x)
On the XY plan, the matching values of y are "under" the draw of y = f(x).

To know where is the concerned region, u can use 1 value of x : x=0 (if definied on 0), calculate f(0) and then look where is a value of y such that y < f(0). U will be "under" the f(x).

Ex 4 : For the line f(x) = -x+1, y < f(x) is in green (fig 4). f(0) = 1 so u know that the point (0,0) is in good region.

Ex 5 : For the line f(x) = 2*x^2 +4*x -1, y < f(x) is in green (fig 5). f(0) = -1 so u know that the point (0,-2) is in good region.
Attachments

Fig1_y_above_3x+1.gif
Fig1_y_above_3x+1.gif [ 2.88 KiB | Viewed 13617 times ]

Fig2_y_above_-2x-1.gif
Fig2_y_above_-2x-1.gif [ 2.76 KiB | Viewed 13612 times ]

Fig3_Y_above_Xpow2-2X-1.gif
Fig3_Y_above_Xpow2-2X-1.gif [ 2.66 KiB | Viewed 13611 times ]

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 [#permalink] New post 16 Jan 2007, 15:19
Fig 4 and 5 :)
Attachments

Fig4_Y_under_-X+1.gif
Fig4_Y_under_-X+1.gif [ 2.73 KiB | Viewed 13585 times ]

Fig5_Y_under_2Xpow2+4X-1.gif
Fig5_Y_under_2Xpow2+4X-1.gif [ 2.6 KiB | Viewed 13593 times ]

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 [#permalink] New post 16 Jan 2007, 15:37
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hey things like distance between (a,b) (c,d) : its a right angle triangle. simple pythagoras formula that is applied there.

all who gave gmat already, Did you try to remember all of these ? I am trying to rely on doing these right there. am i being over-ambitious ?
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 [#permalink] New post 16 Jan 2007, 19:35
Thanks Fig for taking pains to explain. Now, I understand better.
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 [#permalink] New post 13 Jul 2007, 12:29
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wow.... all tis stuff is great... made my day.......... hopefully also my GMAT


Better to understand a little than to misunderstand a lot
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 [#permalink] New post 14 Jul 2007, 12:22
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Thanks for putting in the time to provide this guide, this will be very helpful with my studies!
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Re: Understanding Coordinate Geometry [#permalink] New post 09 Dec 2008, 22:58
thanks for the collection of formulas... I've been looking for something like this.
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Re: Understanding Coordinate Geometry [#permalink] New post 22 Jul 2009, 21:15
Thanks for that one!
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Re: Understanding Coordinate Geometry [#permalink] New post 22 Jul 2009, 22:53
Thanks for collating it at one place.
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Re: Understanding Coordinate Geometry [#permalink] New post 14 Aug 2009, 03:07
thanks for the approach.However,Can somebody please explain me how to plot,
y>f(x),where f(x) = x^2-2x -1.

Thanks in advance.
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Re: Understanding Coordinate Geometry [#permalink] New post 25 Jul 2013, 01:16
Expert's post
amit2k9 wrote:
thanks for the approach.However,Can somebody please explain me how to plot,
y>f(x),where f(x) = x^2-2x -1.

Thanks in advance.


Plot y=x^2-2x-1. y>x^2-2x-1 would be above the graph:
Attachment:
MSP5724200423hdf0492bch00005382e4a5b3eah3fg.gif
MSP5724200423hdf0492bch00005382e4a5b3eah3fg.gif [ 3.1 KiB | Viewed 1268 times ]
Does this make sense?
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Re: I would like to coin together certain properties of [#permalink] New post 27 Jul 2014, 02:24
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Re: I would like to coin together certain properties of   [#permalink] 27 Jul 2014, 02:24
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