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).

Hope it helps.
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IMO A...

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.

(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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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

I am a little confused on this one ...
Here the slope will be negative and line will not pass through Y axis

Line extends in both directions without end (infinitely). What you've drawn there is a line segment.
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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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Target question: Does line k intersect quadrant II?

Statement 1: The slope of k is -1/6
Here are a few lines with slope -1/6

KEY CONCEPT: As we travel from right to left along a line with slope -1/6, we keep moving up.
So, at some point, the line will surely pass through quadrant II
So, the answer to the target question is YES, line k DOES intersect quadrant II
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The y-intercept of k is -6
There are several possible cases that satisfy statement 2. Here are two:
Case a: The line is horizontal (i.e., has slope zero)

In this case, the answer to the target question is NO, line k does NOT intersect quadrant II

Case b: The line looks like this

In this case, the answer to the target question is YES, line k DOES intersect quadrant II

Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
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To visualize:


y=-1x, x int =0, y int =0
y=-1x+3, x int =3, y int =3
y=-1x-3, x int =-3, y int =-3


y=1x, x int =0, y int =0
y=1x+2, x int =-2, y int =2
y=1x-2, x int =2, y int =-2


y=0x+2, x int =none, y int =2
y=0x-2, x int =none, y int =-2

In the rectangular coordinate system shown above, does the line k (not shown) intersect quadrant II?

(1) The slope of k is -1/6
When a slope of a line is negative that means the line is decreasing, and goes downwards from left to right.
So when we extend the line, it will definitely pass through quadrant II and IV.
Hence Statement 1 alone is sufficient.

(2) The y-intercept of k is -6

The y-intercept is the point where the line intersects the y-axis. Here the y intercept of the line is (0,-6).

Case 1:
At (0,-6), we can draw a line parallel to X axis (Slope is 0) and passing through quadrant III and IV. In this case, line doesn't intersect quadrant II

Case 2 :Also we can draw a line passing through (0,-6) with a negative slope passing through the quadrants II,III and IV.
Here, the line will pass through quadrant II

Case 3:
Similarly a line with positive slope can be drawn at the same point passing through the quadrants I,III and IV. Here the line will not intersect quadrant II

Hence we cannot confirm whether the line k intersect quadrant II or not as both cases are possible .

Statement 2 alone is insufficient.

Option A is the answer

Hope this helps,
Clifin J Francis,
GMAT SME
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