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1. If the slope of line is negative, 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 positives, line intersects the quadrant I too, 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. Therefor 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) crosses three quadrants. Only the line which crosses origin (0,0) OR is parallel of either of axis crosses two quadrants.

4. The line with slope 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.

If you draw the couple of lines with different slopes you'll understand this better.

BACK TO THE ORIGINAL QUESTION: In the rectangular coordinate system shown above, does the line k (not shown) intersect quadrant II?

(1) The slope of k is -1/6 (2) The y-intercept of k is -6

Statement (1) says that slope is negative (case 1) so the line will intersect the quadrants II and IV (line goes from up to down). Sufficient.

Statement (2) provides us with y-intercept -6, now if line has positive slope then the line goes from down to up and thus won't intersect quadrant II but if the slope is negative then the line goes from up to down thus it will intersect quadrant II. Two different answers. Not sufficient.

Answer: A.

For more on this topic check Coordinate Geometry chapter of Math Book (link in my signature).

Can someone detail this answer for me? Does any line with a negative slope pass through the second quadrant? Is it because you can put any value in for X, get the value for Y and the line could extend forever?

Yep you're absolutely right. In the gradient intercept formula y=mx+b - The sign of m tells you if its going through 2nd/4th (negative slope) or 1st/3rd (positive slope). The exception to this is if m=0 this will be a horizontal line at b, or if y=0 this will be a vertical line at -b/m.

For the above question this info is sufficient to answer A.

The other info that isn't relevant to this question is. - The value of m (as opposed to the sign) essentially dictates the "steepness" of the line. - The value of B is the Y intercept.

Can someone detail this answer for me? Does any line with a negative slope pass through the second quadrant? Is it because you can put any value in for X, get the value for Y and the line could extend forever?

Yep you're absolutely right. In the gradient intercept formula y=mx+b - The sign of m tells you if its going through 2nd/4th (negative slope) or 1st/3rd (positive slope). The exception to this is if m=0 this will be a horizontal line at b, or if y=0 this will be a vertical line at -b/m.

For the above question this info is sufficient to answer A.

The other info that isn't relevant to this question is. - The value of m (as opposed to the sign) essentially dictates the "steepness" of the line. - The value of B is the Y intercept.

Yes, the question can be answered in about 5 seconds simply by using your logic above:

a. any negative sloping line will cross quadrant 2 at some point because the lines are infinite b. we know nothing about the slope. the slope can be positive or negative. we only know where it crosses

in other words, all you have to do is visualize positive and negative slope

S1: Gives the slope to be -ve and hence it would be as shown in the diagram attached.

S2: Not necessary. without the slope...it is not possible to determine the direction and hence the intersection with the Quandrant.

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Answer to this question is A indeed, as jeeteshsingh explained. The line with negative slope WILL ALWAYS intersect quadrant II and IV. You can also check the Coordinate Geometry chapter of Math Book (link below) for more about this issue.

Statement 1: If the slope is negative then the line will pass through the IInd and IVth quadrant always. Sufficient. Statement 2: Only the y intercept is given, without knowing whether the slope is +ve or -ve, we cannot say whether the line will pass through the II nd quadrant. Insufficient.

(1) Slope is -1/6. So line is y = -x/6 + c. Let x=-6+6c-6|c|. Notice this value of x is negative for all choices of c. If c<0, then x becomes -6+12c, if c=0, then x becomes -6, if c>0, then x becomes -6. At this point, the value of y is 1+|c| which is always positive. Showing that the line will always pass thru the 2nd quadrant (As pointed out earlier this is true for all lines with negative slope) So (1) is sufficient

(2) Y-Int is -6 Consider the line x=-6 (does not pass thru quadrant II) But we know if the line had negative slope it wud pass thru second qudrant, Eg. y=-x-6 Not sufficient

A alone is enough Hence, y-int is not enough to say anything
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Are you sure that the answer is A? The question is confusing in that is the line k an infinite line or a finite line (that is does it have finite ends)?

1. A line with m=-1/6 can lie in I, II and III without intersecting II. It may lie in II as well. Insuff. If unbounded line, this is sufficient. 2. c=-6. Insuff. Clearly a line with positive slope or negative slope could lie in I quadrant. It may intersect II quadrant.

Together, you get an equation y=-x/3 - 6. X and Y intercepts are y=-6, x=-36. This line definitely passes through III quadrant. But, does it have to pass through II.

Depending on if its a bounded or unbounded line, answer could be A or C.
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DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

We are asked if the line passes through the top left quadrant.

1)Slope is negative. A line with a negative slope will definitely pass through the 2nd and 4th quadrants. You can think about it graphically. Unless the line is perfectly vertical, perfectly horizontal or slanting upwards to the right, it will definitely pass through the 2nd and 4th quadrants. Sufficient.

2)As shown below, the line could be one like the green or one like the red. We cant deduce anything from this. Insufficient.

Answer is A

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Re: In the rectangular coordinate system shown above, does the [#permalink]

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08 Feb 2013, 06:31

HI One basic doubt to Statement 2 of the question..... When it is given that Y intercept is -6 , than is it necessary that line must be passing through the point 0,-6 or it can be inferred that its line segment which when extended will pass through 0,-6

In such DS question what should we infer from statement 2 ....Is it a line which for sure is intersecting at point 0,-6 or is just a line segment... Little bit confused......

HI One basic doubt to Statement 2 of the question..... When it is given that Y intercept is -6 , than is it necessary that line must be passing through the point 0,-6 or it can be inferred that its line segment which when extended will pass through 0,-6

In such DS question what should we infer from statement 2 ....Is it a line which for sure is intersecting at point 0,-6 or is just a line segment... Little bit confused......

regards Archit

Strange question... The stem talks about line k and the second statement says that the y-intercept of line k is -6. So, yes, line k passes through point (0, -6).
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In the rectangular coordinate system shown above, does the [#permalink]

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26 Jan 2017, 20:32

Knowing below rules first and attacking problem will save lot of time.

1) If a line has a positive slope, it definitely goes through sectors 1 and 3.

2) If a line has a negative slope, it definitely goes through sectors 2 and 4.

As an aside:

If a line is parallel to the x or y axis, it will go through exactly 2 different sectors.

(Except, of course, for the lines x=0 and y=0, which are the axes.)

If a line is not parallel to the x or y axis it will:

a) go through exactly 2 sectors if it passes through the origin; or

b) go through exactly 3 sectors if it does not pass through the origin.

Now, coming back to statements :-

1) The slope of the line is -(1/6)

You could use trial and error to see that a line with this slope will, eventually, go through sector 2. Sufficient.

(2) the y-int is -6

This tells us that the line passes through (0,-6). We could draw a horizontal line through that point which never touches quadrant 2. We can also draw a diagonal line that does go through quadrant 2. Since the answer is "maybe", (2) is insufficient. Hence A.
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