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# Math: Coordinate Geometry

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Math Expert
Joined: 02 Sep 2009
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09 Feb 2013, 04:53
Archit143 wrote:
HI Bunuel,
As karishma has also mentioned that a line with -ve slope can also be present in 1st quadrant than how can we be so sure that it is intersecting in 2nd quadrant,

Moreover as per the theory it must intersect in 2nd and 4th quadrant ...than as per statement 1 there are two possibilities.....line intersecting in in quad 2 and 4.....But we must have only one answer form the statement, for it to be correct answer..................

Archit

No, that's not what she said.

If the slope of a line is negative, line WILL intersect quadrants II and IV in ANY case. If X and Y intersects are positives, line ALSO intersects the quadrant I, if negative line ALSO intersects the quadrant quadrant III.
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09 Feb 2013, 23:53
Archit143 wrote:
So for a question where it is given that slope is -1/6, Than how can be sure that line intersects 2nd quad, I found this question on GMAT prep......
in-the-rectangular-coordinate-system-shown-above-does-the-90635.html

Archit

Since the slope is negative, the line will intersect the 2nd and 4th quadrant. We are talking about a line, not a line segment. A line extends indefinitely on both ends. The top end of the line will extend to intersect the 2nd quadrant under all circumstances.

Attachment:

Ques3.jpg [ 7.76 KiB | Viewed 4718 times ]

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Karishma
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews VP Status: Final Lap Up!!! Affiliations: NYK Line Joined: 21 Sep 2012 Posts: 1026 Location: India GMAT 1: 410 Q35 V11 GMAT 2: 530 Q44 V20 GMAT 3: 630 Q45 V31 GPA: 3.84 WE: Engineering (Transportation) Re: Math: Coordinate Geometry [#permalink] ### Show Tags 10 Feb 2013, 05:16 Karishma, thats what i wanted to ask....In question it asks about whether the line is intersecting 2nd quadrant....Answer is Yes it does, but at the same time it may lie in 1st quadrant also as explained by you....I think i am badly confused on this.... Archit Math Expert Joined: 02 Sep 2009 Posts: 46251 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 10 Feb 2013, 05:23 Archit143 wrote: Karishma, thats what i wanted to ask....In question it asks about whether the line is intersecting 2nd quadrant....Answer is Yes it does, but at the same time it may lie in 1st quadrant also as explained by you....I think i am badly confused on this.... Archit Does it matter whether the line also lies in other quadrants? We know that it goes through the II and IV quadrants, it may also (simultaneously) go through either I or III quadrant, but this does not alter the fact that the line goes through the II, is it? So, the answer is YES. _________________ Intern Joined: 12 Dec 2016 Posts: 10 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 22 Mar 2017, 13:14 Can you please explain here; To answer, we must find the slope of each line and then check to see if one slope is the negative reciprocal of the other or if their product equals to -1. SlopeAB=5−199−48=−14−39=0.358SlopeAB=5−199−48=−14−39=0.358 SlopeCD=24−422−31=20−9=−2.22 The formula of the slope of two given coordinates are y2-y1 / x2-x1 However in some questions, 2nd coordinates (x2 y2)s are subtracted from 1sts (x1 y1) and in some, other way around. Can you please clarify what do we take into account concerning this formula? Board of Directors Status: Stepping into my 10 years long dream Joined: 18 Jul 2015 Posts: 3651 Re: Math: Coordinate Geometry [#permalink] ### Show Tags 22 Mar 2017, 23:52 1 mesutthefail wrote: Can you please explain here; To answer, we must find the slope of each line and then check to see if one slope is the negative reciprocal of the other or if their product equals to -1. SlopeAB=5−199−48=−14−39=0.358SlopeAB=5−199−48=−14−39=0.358 SlopeCD=24−422−31=20−9=−2.22 The formula of the slope of two given coordinates are y2-y1 / x2-x1 However in some questions, 2nd coordinates (x2 y2)s are subtracted from 1sts (x1 y1) and in some, other way around. Can you please clarify what do we take into account concerning this formula? Hi, This is a mathematical rule. You can write a-b = -(b-a). Did you get this rule?? Now, in a similar way if you take - sign common from both numerator and denominator of y2-y1 / x2-x1, you will get -(y1-y2)/-(x1-x2). I hope you are aware of the rule that - signs can be cancelled out both a numerator and denominator. So, we will be left with the formula, Slope = (y1-y2)/(x1-x2). I hope that makes sense. _________________ My GMAT Story: From V21 to V40 My MBA Journey: My 10 years long MBA Dream My Secret Hacks: Best way to use GMATClub Verbal Resources: All SC Resources at one place | All CR Resources at one place GMAT Club Inbuilt Error Log Functionality - View More. NEW VISA FORUM - Ask all your Visa Related Questions - here. How to use an Error Log? Check out my tips! Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8102 Location: Pune, India Re: Math: Coordinate Geometry [#permalink] ### Show Tags 23 Mar 2017, 04:22 mesutthefail wrote: Can you please explain here; To answer, we must find the slope of each line and then check to see if one slope is the negative reciprocal of the other or if their product equals to -1. SlopeAB=5−199−48=−14−39=0.358SlopeAB=5−199−48=−14−39=0.358 SlopeCD=24−422−31=20−9=−2.22 The formula of the slope of two given coordinates are y2-y1 / x2-x1 However in some questions, 2nd coordinates (x2 y2)s are subtracted from 1sts (x1 y1) and in some, other way around. Can you please clarify what do we take into account concerning this formula? They are both the same. $$Slope = \frac{(y2 - y1)}{(x2 - x1)} = \frac{(y1 - y2)}{(x1 - x2)}$$ Take an example: (x1, y1) = (2, 3) (x2, y2) = (5, -10) $$Slope = \frac{(y2 - y1)}{(x2 - x1)} = \frac{-10 - 3}{5 - 2} = -\frac{13}{3}$$ $$Slope = \frac{(y1 - y2)}{(x1 - x2)} = \frac{3- (-10)}{2 - 5} = -\frac{13}{3}$$ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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02 Sep 2017, 00:22
Hello Bunuel,

Thanks a lot for the article. I have one doubt . Is it that a line with negative slope would definitely pass through Quadrant 2 and 4, and would pass through 1 or 3 depending on the value of x and y intersects?
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02 Sep 2017, 02:57
Arsh4MBA wrote:
Hello Bunuel,

Thanks a lot for the article. I have one doubt . Is it that a line with negative slope would definitely pass through Quadrant 2 and 4, and would pass through 1 or 3 depending on the value of x and y intersects?

Yes. If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.
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06 Sep 2017, 18:11
Hi Bunuel,

Under heading Line equation in Co-ordinate Geometry -

The equation of a straight line passing through points P1(x1,y1)P1(x1,y1) and P2(x2,y2)P2(x2,y2) is:

$$\frac{y−y1}{x−x1}=\frac{y1−y2}{x1−x2}$$

I think it should be :

$$\frac{y−y1}{x−x1}=\frac{y2−y1}{x2−x1}$$

This is because slope for two points is : $$\frac{y2-y1}{x2-x1}$$

Let me know if I am missing anything here.
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06 Sep 2017, 21:34
1
BloomingLotus wrote:
Hi Bunuel,

Under heading Line equation in Co-ordinate Geometry -

The equation of a straight line passing through points P1(x1,y1)P1(x1,y1) and P2(x2,y2)P2(x2,y2) is:

$$\frac{y−y1}{x−x1}=\frac{y1−y2}{x1−x2}$$

I think it should be :

$$\frac{y−y1}{x−x1}=\frac{y2−y1}{x2−x1}$$

This is because slope for two points is : $$\frac{y2-y1}{x2-x1}$$

Let me know if I am missing anything here.

Both are the same: $$\frac{y2−y1}{x2−x1}=\frac{-(y1−y2)}{-(x1−x2)}=\frac{y1-y2}{x1-x2}$$
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12 Sep 2017, 17:37
Can anyone guide me where i can practice only coordinate plane questions??
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12 Sep 2017, 21:00
1
SandhyAvinash wrote:
Can anyone guide me where i can practice only coordinate plane questions??

Use our search engine to find questions from specific category: https://gmatclub.com/forum/search.php?view=search_tags
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18 Sep 2017, 22:10
Bunuel sir, I need coordinate Geometry in PDF. I want to download the attachment
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18 Sep 2017, 23:26
Raj94* wrote:
Bunuel sir, I need coordinate Geometry in PDF. I want to download the attachment

Check here: https://gmatclub.com/forum/bunuel-signa ... 70062.html
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Re: Math: Coordinate Geometry   [#permalink] 18 Sep 2017, 23:26

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