Hi, there.

With all due respect to pemdas, I'd like to put in my 2¢ here.

Prompt: does the Line k (not shown) intersect quadrant II?

Statement #1: The slope of k is -1/6

A slope = -1/6 is a negative slope, a downward slope that goes up on the left and down on the right. Since a line is infinite, any line going down to the right is going to hit Quadrant IV at some point. Similarly, any line going up to the left is going to hit Quadrant II at some point. A line with a negative slope may or may not pass through Quadrants I or III, and categorically cannot pass through both of them, but a line with a negative slope must pass through both Quadrant II and Quadrant IV. Thus, Statement #1, by itself, is sufficient.

Statement #2: The y-intercept of k is -6

Well, this means . . .
(a) if the line has a positive slope, it goes through III, IV, and I
(b) if the line has a negative slope, it goes through II, III, and IV
(c) if the line is horizontal, it goes through III and IV only

Since with Statement #2, by itself, we have no information about the slope, then this statement by itself is insufficient.

Answer = A

Here's another coordinate geometry DS question for practice.

https://gmat.magoosh.com/questions/1031

Does everything I've said make sense? If you have any further questions, please do not hesitate to ask.

Mike

pappueshwar wrote:hi Mike,

request to explain this slope concept and y intercept concepts once again in regards with movement up , down and side ways... i am of the opinion that the slope should be read from the perspective of Y intercept

say the equation given is y = 1/2x - 2
so slope here is 1/2
y intercept is -2.

so when we want to plot this on the co-ordinate plane first place a point on -2, from there on since it is a positive slope, where in change in Y is 1 and change in X is 2 so u need to move 1 point up and 2 points right so the point is (2,-1) correct

in this case:

slope is -1/6
y intercept -6

so place the point on -6 (y-intercept), since the change in y is -1 the point moves down and moves right for change in X to 6 places making the point (6, -7).correct

let me know on this. i know i am wrong but i wanna get the concept right...
similar concept can be applied here and hence i arrived the answer as C. not correct

Dear pappueshwar

You do understand the basic concept of slope quite well. The only thing I'll add, which may be obvious to you, is: a slope of 1/2 can mean move right 2 and up 1, but it can equally well mean move left 2 and down 1. Starting with a slope of 1/2 and a y-intercept of -2, moving to the right, we get:

(0, -2) --> (2, -1) --> (4, 0) --> (6, 1) etc.

and moving left, we get:

(0, -2) --> (-2, -3) --> (-4, -4) --> (-6, -5) etc.

I assume those ideas about slope also make sense.

Understanding slope itself is not a problem here. Let's talk about this particular question. Again, the 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.

This is a tricky question.

Statement #1: We know the line has a negative slope, a slope of -1/6. We don't know the y-intercept. We want to know: does the line pass through QII? Well, if the y-intercept is a positive number, then the line definitely passes through QII. What if the y-intercept is negative? If the y-intercept is negative, the line definitely goes through QIII and QIV, but what about QI or QII?

Well, if the slope is negative, it goes down to the right and up to the left. In particular, as we go an infinite distance to the left, the line continues to rise without limit. That means, at some point, it must enter QII.

Another way to say it: every non-horizontal line has an x-intercept. Line k has slope of m = -1/6, so it definitely has to have an x-intercept. If the y-intercept is negative and the slope is negative, that means the x-intercept cannot occur on the positive x-axis --- it must occur on the negative x-axis. See diagram in the attached pdf. If line k has an x-intercept on the negative x-axis, that means it must enter QII.

Thus, statement #1, by itself, is sufficient.

As I think you appreciated, statement #2, by itself, is insufficient, so the correct answer is A.

TAKEAWAY:
1) Any line with a positive slope, regardless of y-intercept, must go through QI and QIII
2) Any line with a negative slope, regardless of y-intercept, must go through QII and QIV

Does all this make sense? Please let me know if you have any further questions.

Mike

pappueshwar wrote:mike,

thanks.

can u explain statement 2 also ...

Dear pappueshwar,

I've attached a diagram with two lines in the x-y plane. Both have a y-intercept of -6. Line #1 has a positive slope, and line #2 has a negative slope. As we can see from the diagram, any line with a negative y-intercept and a positive slope will go through Q3, Q4, and Q1. Any line with a negative y-intercept and a negative slope will go through Q2, Q3, and Q4.

The question this DS addresses is: does the line go through Q2?

Statement #2 tells us only that the y-intercept is -6. That is not enough to determined whether it goes through Q2. We would need to know the slope --- with a negative slope, it would go through Q2, and with a positive slope, it wouldn't. Because we don't know, this statement is insufficient.

As discussed above, statement #1 is sufficient.

Therefore, answer = A.

Does that make sense? Please let me know if you have any further questions.

Mike