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How to plot the following in coordinate system

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by goelmohit2002 » Wed Sep 16, 2009 4:52 am
Hi All,

Can someone please tell how to plot the following equation area in coordinate system and what is the reasoning for the same ?

2x + 3y <= 6

I was able to draw the line that represent 2x + 3y = 6. But how to draw the area that represents 2x + 3y <= 6.

Thanks
Mohit
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by kyabe » Wed Sep 16, 2009 5:10 am
The simplest way is to plug-in values from both the sides of line and whichever satisfies the equation will be ur side.

Tip: If the line is not passing through (0,0) use this co-ordinate to decide.
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by goelmohit2002 » Wed Sep 16, 2009 5:34 am
kyabe wrote:The simplest way is to plug-in values from both the sides of line and whichever satisfies the equation will be ur side.

Tip: If the line is not passing through (0,0) use this co-ordinate to decide.
Thanks Kyabe.

plugging (0,0)...yields

0 <= 6...which is not true...

Can you please tell does it mean that all the points below the line will not satisfy this inequality....i.e. all the points above the line satisfy this equation ?

Thanks
Mohit
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by kyabe » Wed Sep 16, 2009 5:36 am
goelmohit2002 wrote:
kyabe wrote:
plugging (0,0)...yields

0 <= 6...which is not true...

Can you please tell does it mean that all the points below the line will not satisfy this inequality....i.e. all the points above the line satisfy this equation ?

Thanks
Mohit
Are u sure its not true???


For me it is true and hence the side containing 0,0 is the side satisfying the equation
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by goelmohit2002 » Wed Sep 16, 2009 5:41 am
kyabe wrote:
goelmohit2002 wrote:
kyabe wrote:
plugging (0,0)...yields

0 <= 6...which is not true...

Can you please tell does it mean that all the points below the line will not satisfy this inequality....i.e. all the points above the line satisfy this equation ?

Thanks
Mohit
Are u sure its not true???


For me it is true and hence the side containing 0,0 is the side satisfying the equation
sorry sorry my mistake....yes 0 <=6 :-)

so in these type of questions....does it mean that all the points on one side of line are less than and all other points on other side are greater than ?

Any mathematical reason for the same ?
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by kyabe » Wed Sep 16, 2009 5:43 am
Try a few and check urself :) .. If u want the proof of it u need to contact a mathematics professor..
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by goelmohit2002 » Wed Sep 16, 2009 5:48 am
kyabe wrote:Try a few and check urself :) .. If u want the proof of it u need to contact a mathematics professor..
Thanks Kyabe....

Actually what I meant was not proof....but reason behind this trick ?..so that we can be sure that this trick work all the time and has no dreaded exception.
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by kyabe » Wed Sep 16, 2009 5:53 am
I dont know how to answer your question.. Only thing I can assure you that this way u'll never get wrong answer..
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by goelmohit2002 » Wed Sep 16, 2009 5:54 am
kyabe wrote:I dont know how to answer your question.. Only thing I can assure you that this way u'll never get wrong answer..
Thanks a lot Kyabe...that helps !!!
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by goelmohit2002 » Wed Sep 16, 2009 8:05 am
Hi Kyabe,

Following is the related question from OG-12, Q121. Can you please tell how to solve the same without using number picking....OG has used some random set of numbers to prove the same....but finding those random set is I think very dificult.

In the xy plane, region R consists of all the points (x, y) such that 2x + 3y <= 6. Is the point (r,s) in the Region R ?
1) 3r + 2s = 6
2) r <= 3 and s <= 2

[spoiler]OA = E.[/spoiler]
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by kyabe » Wed Sep 16, 2009 8:46 am
See the attached document.. Its the most simple way to solve problems like this...


The area marked in RED is not covered by the line 2x + 3y < 6. So the answer is E
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by goelmohit2002 » Wed Sep 16, 2009 8:52 am
Aweome Kyabe ! Fantastic !

You are really ahead of OG....

I guess this is the reason why GMAT gives graph type paper in exam. First question where I have encountered the use of graph....without graph this would have been almost impossible to correctly make.

What do you say ?
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by kyabe » Wed Sep 16, 2009 8:56 am
I disagree to the statment "You are really ahead of OG.... "

Partially agree to "without graph this would have been almost impossible to correctly make" :wink:
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