## PS- Coordinate Geometry

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### PS- Coordinate Geometry

by pareekbharat86 » Mon Nov 04, 2013 10:11 pm
In the rectangular coordinate system which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x âˆ’ 3y â‰¤ âˆ’ 6 ?
(A) None
(B) Î™
(C) Î™I
(D) Î™II
(E) IV

OA is E.

My Response to the ques:

Assuming x=0, then -3y<= -6 or y>= 2.

Assuming y=0, x<= -3.

Therefore a line with coordinates (X<= -3, Y>=2) should be located in quadrant 2 only.

Having done this much, I got stuck. I did not really comprehend what the question stem meant.
Thanks,
Bharat.

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by theCodeToGMAT » Mon Nov 04, 2013 10:51 pm
2x - 3y <= -6

x = 0==> -3y <= -6 ==> y>=2

y = 0==> 2x <= -6 ==> x <= -3

Since, any line drawn using the above points will result lines passing 1st, 2nd & 3rd quadrant.
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by mevicks » Mon Nov 04, 2013 11:04 pm
pareekbharat86 wrote:In the rectangular coordinate system which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x âˆ’ 3y â‰¤ âˆ’ 6 ?
(A) None
(B) Î™
(C) Î™I
(D) Î™II
(E) IV
Let us first trace the region depicted by 2x âˆ’ 3y â‰¤ âˆ’ 6
2x âˆ’ 3y â‰¤ âˆ’ 6
âˆ’3y â‰¤ -2x âˆ’ 6
3y â‰¥ 2x + 6 ... multiplying both sides by (-1)

y â‰¥ (2/3)x + 2

Consider the line y = (2/3)x + 2
Its y intercept = 2 (put x = 0 in the above equation)
Its x intercept = -3 (put y = 0 in the above equation)

We need a region which is greater than y = (2/3)x + 2
(since the original equation states greater than equal to...) Original question: which quadrant contains no point that satisfies the inequality ...

The points in the 4th quadrant will not be a part of the equation y â‰¥ (2/3)x + 2 and thus the answer is E

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by [email protected] » Tue Nov 05, 2013 4:56 am
pareekbharat86 wrote:In the rectangular coordinate system which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x âˆ’ 3y â‰¤ âˆ’6 ?
(A) None
(B) Î™
(C) Î™I
(D) Î™II
(E) IV
For those who are unfamiliar with the location of the quadrants in the x-y coordinate plane, they are shown here Another approach here is the recognize how the coordinates of points look in each quadrant.

From here, we can check each quadrant.
For example, let's see if it's possible for a point in QUADRANT I (where the x and y coordinates are both positive) to satisfy the inequality 2x âˆ’ 3y â‰¤ âˆ’6
Well, how about x = 1 and y = 10.
When we plug these in, we get 2(1) âˆ’ 3(10) â‰¤ âˆ’6
Simplify to get -28 â‰¤ âˆ’6 [PERFECT it works]
ELIMINATE B

Is it possible for a point in QUADRANT II (where the x is negative and y is positive) to satisfy the inequality 2x âˆ’ 3y â‰¤ âˆ’6?
Well, how about x = -10 and y = 10.
When we plug these in, we get 2(-10) âˆ’ 3(10) â‰¤ âˆ’6
Simplify to get -50 â‰¤ âˆ’6 [PERFECT it works]
ELIMINATE C

Is it possible for a point in QUADRANT III (where the x is negative and y is negative) to satisfy the inequality 2x âˆ’ 3y â‰¤ âˆ’6?
Well, how about x = -10 and y = -1.
When we plug these in, we get 2(-10) âˆ’ 3(-1) â‰¤ âˆ’6
Simplify to get -17 â‰¤ âˆ’6 [PERFECT it works]
ELIMINATE D

Is it possible for a point in QUADRANT IV (where the x is positive and y is negative) to satisfy the inequality 2x âˆ’ 3y â‰¤ âˆ’6?
NO.
When we plug these in, we get 2(positive) âˆ’ 3(negative) â‰¤ âˆ’6
Simplify to get positive - negative â‰¤ âˆ’6
Since a positive number minus a negative number will always be positive, we get . . .
POSITIVE â‰¤ âˆ’6
Since this is IMPOSSIBLE, the correct answer must be E

Cheers,
Brent

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