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I had a question regarding this: 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

Is that right? When you say x intercept is 5 then the points are (0,5) right? Same for y intercept is 2..(2,0). So when you look at the equation in slope intercept form it should be y = (-5/2)x + 5

1. 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.

2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intersects of the line with positive slope have opposite signs. Therefore if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV.

3. Every line (but the one crosses origin OR parallel to X or Y axis OR X and Y axis themselves) crosses three quadrants. Only the line which crosses origin \((0,0)\) OR is parallel to either of axis crosses only two quadrants.

4. If a line is horizontal it has a slope of \(0\), is parallel to X-axis and crosses quadrant I and II if the Y intersect is positive OR quadrants III and IV, if the Y intersect is negative. Equation of such line is y=b, where b is y intersect.

5. If a line is vertical, the slope is not defined, line is parallel to Y-axis and crosses quadrant I and IV, if the X intersect is positive and quadrant II and III, if the X intersect is negative. Equation of such line is \(x=a\), where a is x-intercept.

6. For a line that crosses two points \((x_1,y_1)\) and \((x_2,y_2)\), slope \(m=\frac{y_2-y_1}{x_2-x_1}\)

7. If the slope is 1 the angle formed by the line is \(45\) degrees.

8. Given a point and slope, equation of a line can be found. The equation of a straight line that passes through a point \((x_1, y_1)\) with a slope \(m\) is: \(y - y_1 = m(x - x_1)\)

I was going through "GMAT MATH BOOK" for Coordinate Geometry. Came across a section(SLOPE AND QUADRANTS:) that deals with intersection of lines with Quadrants. This got me confused as it does not match with my concepts that i know of.

According to me (atleast what i followed till now) , when we talk about intersection of a line with Quadrant. I used to think that we are dealing with the quadrant with which the line forms a triangle.(using axes as other two sides) This does not match with the above mentioned topic.

Can ne one help me with this and a few GMAT problems that deal with this would certainly help.

I just wanted to clear my understanding of certain concepts.

Re: Math: Coordinate Geometry [#permalink]
08 Dec 2011, 11:02

Good work. Really helpful.

Q: If the point (x,y) moves such that the sum of its distances from the points (0,2) and (0,-2) is 6, then the point (x,y) lies on : 1. 9x+5y + 45 2. 9x^2 + 5y^2 = 45 3. 9x+ 5y^2 = 45

Can you please answer the above question. _________________

Re: Math: Coordinate Geometry [#permalink]
08 Jan 2012, 13:42

Can some one validate the statement below

1. 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.

2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intersects of the line with positive slope have opposite signs. Therefore if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV.

If the point is in the 4th quadrant , x and y will have different sign. But statement 1 contradicts this belief.

If the point is in 1st or 3rd quadrant, x and y will have same sign. But the statement 2 above contradicts this belief.

Can some one help me to understand if im wrong? Vidhya

Re: Math: Coordinate Geometry [#permalink]
06 Mar 2012, 06:25

[quote="Bunuel"]

SLOPE AND QUADRANTS:

1. 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.

2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intersects of the line with positive slope have opposite signs. Therefore if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV.

Could someone please explain this further. What do you mean by "intersect" here. Oh and I just saw the same question asked in the previous post _________________

If you like it, Kudo it!

"There is no alternative to hard work. If you don't do it now, you'll probably have to do it later. If you didn't need it now, you probably did it earlier. But there is no escaping it."

Re: Math: Coordinate Geometry [#permalink]
06 Mar 2012, 07:13

Expert's post

budablasta wrote:

Bunuel wrote:

SLOPE AND QUADRANTS:

1. 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.

2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intersects of the line with positive slope have opposite signs. Therefore if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV.

Could someone please explain this further. What do you mean by "intersect" here. Oh and I just saw the same question asked in the previous post

Intersects mean passes through, have points in common. _________________

really great job, Bunuel!!! i have learnt from this post more than did in school + university ! thank you for help and for your time you spent to create this stuff! Well done!!!sincerely

When I have two points in a coordinate system e.g. (2,3) and (6,7) that pass through a line how do I know which number is x1 and which is x2 when calculating slope. Also, the same for the y1 and y2? _________________

A small remark at "7. If the slope is 1 the angle formed by the line is 45 degrees." Angle formed by the line with who? The slope of a line is a real number equal to the tangent of the angle formed by the line with the positive x axis. For the GMAT, no need to know about tangent (trigonometry), just to understand which angle we are talking about. If the angle is acute, the slope (tangent of the angle) is positive. If the angle is obtuse, the slope (tangent of the angle) is negative. There is a one-to-one correspondence between angles and the real numbers representing their so-called tangent values. So, a line with slope 1 forms an angle of 45 degrees and a line with slope -1 forms an angle of 135 degrees with the positive x axis. If the angle is 60 degrees, the slope is \(\sqrt{3}\), and \(-\sqrt{3}\) if the angle is 120 degrees, etc.

A suggestion: wouldn't be better to define the slope somewhere before the equations? I am sure everybody knows what's the slope of a line, just for the coherence. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Math: Coordinate Geometry [#permalink]
08 Feb 2013, 05:20

HI I have a basic doubt , which deals with concept Previously , I have learnt that a line that has rising slope when moving from - ve X to + ve X will have +ve slope. But I found in the original post of BUNNUEL that a line with - ve slope will always be in 2nd or 4th quadrant. I feel that these two statements are contradictory. We can have a line with + ve slope even if it is in 2nd quadrant. Than how can we come to conclusion that whenever we encounter a slope with - ve sign , thn it must either lie in 2nd or lie in 4 th quadrant. As we can have lines in 1st quadrant with - ve slope also........

Re: Math: Coordinate Geometry [#permalink]
09 Feb 2013, 01:17

Expert's post

Archit143 wrote:

HI I have a basic doubt , which deals with concept Previously , I have learnt that a line that has rising slope when moving from - ve X to + ve X will have +ve slope. But I found in the original post of BUNNUEL that a line with - ve slope will always be in 2nd or 4th quadrant. I feel that these two statements are contradictory. We can have a line with + ve slope even if it is in 2nd quadrant. Than how can we come to conclusion that whenever we encounter a slope with - ve sign , thn it must either lie in 2nd or lie in 4 th quadrant. As we can have lines in 1st quadrant with - ve slope also........

Responding to a pm:

A line with +slope MUST lie in the 1st and 3rd quadrant. It can also lie in either the 2nd or the 4th quadrant or it may not lie in both 2nd and 4th (if it passes through the center). But it must lie in 1st as well as the 3rd quadrant.

A line with -ve slope MUST lie in the 2nd and 4th quadrant. It can also lie in either 1st or 3rd quadrant or it may not lie in both 1st and 3rd (if it passes through the center). But it must lie in 2nd as well as the 4th quadrant.

Draw some lines with +ve/-ve slopes to figure this out. _________________

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