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# Math: Coordinate Geometry

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Veritas Prep GMAT Instructor
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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 8171 times ]

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Karishma
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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
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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.
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22 Mar 2017, 13:14

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?
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22 Mar 2017, 23:52
1
mesutthefail wrote:

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.
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23 Mar 2017, 04:22
mesutthefail wrote:

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)

$$Slope = \frac{(y2 - y1)}{(x2 - x1)} = \frac{-10 - 3}{5 - 2} = -\frac{13}{3}$$

$$Slope = \frac{(y1 - y2)}{(x1 - x2)} = \frac{3- (-10)}{2 - 5} = -\frac{13}{3}$$
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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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22 Aug 2018, 20:40
Bunuel wrote:

4. Intercept form.
The equation of a straight line whose x and y intercepts are a and b, respectively, is:
$$\frac{x}{a}+\frac{y}{b}=1$$

Example #3
Q: Find the equation of a line whose x intercept is 5 and y intercept is 2.
Solution: Substituting the values in equation $$\frac{x}{a}+\frac{y}{b}=1$$ we'll get $$\frac{x}{5}+\frac{y}{2}=1$$ --> $$5y+2x-10=0$$ OR if we want to write the equation in the slope-intercept form: $$y=-\frac{2}{5}x+2$$

I can't figure out how you got the -10 in this 5y+2x-10=0 from $$\frac{x}{5}+\frac{y}{2}=1$$
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22 Aug 2018, 21:31
1
climbguy wrote:
Bunuel wrote:

4. Intercept form.
The equation of a straight line whose x and y intercepts are a and b, respectively, is:
$$\frac{x}{a}+\frac{y}{b}=1$$

Example #3
Q: Find the equation of a line whose x intercept is 5 and y intercept is 2.
Solution: Substituting the values in equation $$\frac{x}{a}+\frac{y}{b}=1$$ we'll get $$\frac{x}{5}+\frac{y}{2}=1$$ --> $$5y+2x-10=0$$ OR if we want to write the equation in the slope-intercept form: $$y=-\frac{2}{5}x+2$$

I can't figure out how you got the -10 in this 5y+2x-10=0 from $$\frac{x}{5}+\frac{y}{2}=1$$

$$\frac{x}{5}+\frac{y}{2}=1$$;

$$\frac{2x+5y}{10}=1$$;

$$2x+5y=10$$;

$$5y+2x-10=0$$.

Hope its clear.
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16 Sep 2018, 00:54
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Dear Bunuel,
Thanks for the wonderful explanation of coordinate geometry. The math guide is a surely a saviour.
I just wanted to bring your attention to the below quote, which is mentioned in the book.

Quote:
A horizontal line has a slope of zero.
The equation of a horizontal line is:
y=b
Where: x is the coordinate of any point on the line; b is where the line crosses the y-axis (y intercept). Notice that the equation is independent of x. Any point on the horizontal line satisfies the equation.

I think the highlighted text in red should be y instead of x.

Thanks
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01 Feb 2019, 03:37
aalekhoza wrote:
Dear Bunuel,
Thanks for the wonderful explanation of coordinate geometry. The math guide is a surely a saviour.
I just wanted to bring your attention to the below quote, which is mentioned in the book.

Quote:
A horizontal line has a slope of zero.
The equation of a horizontal line is:
y=b
Where: x is the coordinate of any point on the line; b is where the line crosses the y-axis (y intercept). Notice that the equation is independent of x. Any point on the horizontal line satisfies the equation.

I think the highlighted text in red should be y instead of x.

Thanks

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Fixed that. Thank you.
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Re: Math: Coordinate Geometry   [#permalink] 01 Feb 2019, 03:37

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