DS: Does k intersect in QII? EDIT: the correct question is "does line k intersect quadrant II?" - no "in"
1) k=-1/6
2)y intercept = -6

In the x-y plane, what is the y-intercept of line l
1) slope of l = 3* y-intercept
2) x-intecept of line l = -1/3

In the first question,
k is the line
1) slope of line k = -1/6

just to clarify

The source of these questions is mba.com review questions

Note: I edited the problems slightly. Also, did you have a specific question? did you get stuck on something? in any case, here are the solutions...

Q1.
DS: Does line k intersect in QII?
1) the slope of line k=-1/6
2)y intercept = -6

The answer is A.


Rephrase for algebra lovers: Does line K have any points (x,y) such that x is negative and y is positive?
Rephrase for geometers: Does line K go into the upper left quadrant?

statement 1
algebra rephrase: y = -1/6x + b
when x is a really big negative number, -1/6x is a really big positive number. hence y will also be positive, and the answer to the question is YES. therefore statement 1 is sufficient.

geometry rehrase: the line slopes up and to the left. so when it goes really far to the left, it goes really high up. therefore it will go to the upper left quadrant and the answer is YES. therefore statement 1 is sufficient.


statement 2
rephrase:
y = mx - 6
if m is positive, then it will not go into the 2nd quadrant, because when x is negative y will also be negative. if m is negative, it will go into the second quadrant. we don't know if m is positive or negative, hence insufficient.

Q2
In the x-y plane, what is the y-intercept of line l
1) slope of l = 3* y-intercept
2) x-intecept of line l = -1/3

The answer is E.

1. the slope can be anything hence the intercept can be anything. insufficient.
(for example, we could have y = 3x + 1 or y = 3,000,000x + 1,000,000)

2. knowing the x-intercept does not tell us the y-intercept because the slope can be anything.

together: now we know one point on the line (x = -1/3, y = 0). that is the x-intercept. now consider the y-intercept, 'b'. the coordingates of this point are (0, b). we don't know the value of b, but we can say the slope would be the change in y over the change in x between the two points. therefore it would be (b - 0)/(0 + 1/3) = b/3. this tells us that the slope MUST be 3 times the y-intercept, NO MATTER WHAT the y-intercept is (and hence no matter what the slope is). hence we cannot find the slope. answer is E.

Hi,

I cannot follow the explanation given above, even though the language above is very easy. I am too weak with coordinate geometry..

Especially "when x is a really big negative number, -1/6x is a really big positive number. hence y will also be positive, and the answer to the question is YES. therefore statement 1 is sufficient"

I do not understand a word of the above statement..

Can someone please explain this sum in a more simple way ? Thanks and appreciate your patience..

Also can you suggest some online material for Coordinate geometry, as advised by the instructors I searched google for around 20 mins but the material I found was too basic. Thanks

Someone please answer this.. Thnx

if the slope is - ve the line has to pas thru the second or fourth quadrant .........
keep this rule in mind
frst has to be A ......

Kamlesh,

Thanks for the response... But my confusion is what if the points intersect only in quadrant 4 and not in 2.. I think I am misinterpreting the question... Also could you kindly suggest some material for coordinate geomtery.. I am too weak with this chapter.. Thanks appreciate ur help

if the slope of the line is -1/6, then the line's equation can be written as y = (-1/6)x + b, where b is the y-intercept.
to ensure the existence of a point in the second quadrant, choose x to be a ginormous negative number, like, say, -600,000,000,000,000.
then y = +100,000,000,000,000 + b, which should definitely be positive. if "b" is such a huge negative number that even this is still negative, then just make x even more negative until that doesn't happen anymore.
eventually, you'll be able to find a point with negative x and positive y, meaning it's in the second quadrant.

--

in any case, ALGEBRA IS NOT AN EFFICIENT WAY TO SOLVE MOST GMAT PROBLEMS ABOUT SLOPES AND QUADRANTS. YOU SHOULD LEARN TO CONCEPTUALIZE SLOPES. in other words, you should internalize the idea of positive slope vs. negative slope vs. zero slope vs. undefined slope, and you should be able to think about what those slopes LOOK like.
if you realize that a slope of -1/6 goes up to the left and down to the right, then it's easy to see that it eventually has to go into the upper left quadrant. compared to that, why would anyone want to bother with all that algebra?

the problem actually lies with the original post; the problem has been mistranscribed.

the real problem asks, "does line k intersect quadrant II?"
the word "in" isn't there; the addition of "in" in the original post makes it nonsensical (a line can't "intersect" in a quadrant - that doesn't mean anything, because one item can't "intersect" by itself).

the question is asking whether the line intersects the quadrant - i.e., whether the line contains any points from the quadrant.

i have edited the original post.