whats that square box with 2??

IMO D

I think the point (s,-2) is like a line parallel to the X axis which runs in the lower two quadrants. And our line k, since it has a negative slope is of the shape \
SO its X intercept could be positive negative or zero or anything.
(1) says that s is 0. This means k passes through Y axis at (0,-2). Obviously its x intercept is negative.
SUFF
(2) says that y intercept of k is negative. Visualize it!!. (s.-2) is one point and te other point is on the negative side of Y axis. So the x intercept has to be negative. Thus this is SUFF too.

umaa wrote:In the xy-plane, if line k has negative slope and passes
through the point (s, -2), is the x-intercept of line k positive?
(1) s = 0
(2) The y-intercept of line k is negative.

Don't know the original answer. IMO E.

Let me know your answers and explanations.

(Edited)

Very tough question to answer in words as opposed to a sketch (which is what you should always draw on your scratch paper for geometry word problems).

Let's break the question down.

Step 1 of the Kaplan Method for DS: Focus on the question stem

We know that our line has a negative slope (goes from top left to bottom right) and passes through the point (s, -2), i.e. it travels below the x-axis. There are three possibilities for the location of that point:

1) (s, -2) could be in the lower left quadrant (-, -). If a line with a negative slope passes through the lower left quadrant, then it's x-intercept will be negative.

2) (s, -2) could be on the y-axis, i.e. (0, -2). If a line with a negative slope has a negative y-intercept, then it's x-intercept will be negative.

3) (s, -2) could be in the lower right quadrant (+, -). If a line with a negative slope passes through the lower right quadrant, then it's x-intercept could be negative, 0 or positive.

So, if we can narrow it down to one of the first two possibilities, we can answer the question. If we can only narrow it down to the 3rd possibility, then without further info we can't answer the question.

Step 2 of the Kaplan Method for DS: consider each statement independently

(1) s = 0

Perfect! Based on our preliminary work, we know we're in case #2 and the x-intercept will definitely be negative: sufficient.

(2) The y-intercept of line k is negative.

Good enough! Identical to our reasoning for case #2 - we have a line with a negative slope and a negative y-intercept, therefore the x-intercept will also be negative: sufficient.

Step 3 of the Kaplan Method for DS: If necessary, combine the statements

No need for step 3 on this question!

Each of (1) and (2) is sufficient alone: choose (D).

Hi Stuart,

Nice explanation. Just one doubt. If the line is in the lower right quandrant i.e. quadrant 4, then y should be -ve. But you've said that:

(s, -2) could be in the lower right quadrant (-, +).

Can you clarify please?

Thanks,
Vikas Mehra

Stuart Kovinsky wrote:

umaa wrote:In the xy-plane, if line k has negative slope and passes
through the point (s, -2), is the x-intercept of line k positive?
(1) s = 0
(2) The y-intercept of line k is negative.

Don't know the original answer. IMO E.

Let me know your answers and explanations.

(Edited)

Very tough question to answer in words as opposed to a sketch (which is what you should always draw on your scratch paper for geometry word problems).

Let's break the question down.

Step 1 of the Kaplan Method for DS: Focus on the question stem

We know that our line has a negative slope (goes from top left to bottom right) and passes through the point (s, -2), i.e. it travels below the x-axis. There are three possibilities for the location of that point:

1) (s, -2) could be in the lower left quadrant (-, -). If a line with a negative slope passes through the lower left quadrant, then it's x-intercept will be negative.

2) (s, -2) could be on the y-axis, i.e. (0, -2). If a line with a negative slope has a negative y-intercept, then it's x-intercept will be negative.

3) (s, -2) could be in the lower right quadrant (-, +). If a line with a negative slope passes through the lower right quadrant, then it's x-intercept could be negative, 0 or positive.

So, if we can narrow it down to one of the first two possibilities, we can answer the question. If we can only narrow it down to the 3rd possibility, then without further info we can't answer the question.

Step 2 of the Kaplan Method for DS: consider each statement independently

(1) s = 0

Perfect! Based on our preliminary work, we know we're in case #2 and the x-intercept will definitely be negative: sufficient.

(2) The y-intercept of line k is negative.

Good enough! Identical to our reasoning for case #2 - we have a line with a negative slope and a negative y-intercept, therefore the x-intercept will also be negative: sufficient.

Step 3 of the Kaplan Method for DS: If necessary, combine the statements

No need for step 3 on this question!

Each of (1) and (2) is sufficient alone: choose (D).

mehravikas wrote:Hi Stuart,

Nice explanation. Just one doubt. If the line is in the lower right quandrant i.e. quadrant 4, then y should be -ve. But you've said that:

(s, -2) could be in the lower right quadrant (-, +).

Can you clarify please?

Thanks,
Vikas Mehra

Sorry, typo on my part - lower right is (+, -).

Thanks, I edited my post accordingly.

Stuart, Thanks a lot for the great explanation. I didn't know the above three cases about negative slopes.

Staurt,
I created tables for both positive and negative slopes for easy access. Can you please tell me whether my tables are correct?

D

REALLY A GOOD ONE

U CAN USE

SLOPE= - YINT\ X INT

AS SLOPE IS -VE

TWO SITUATIONS

1 BOTH XINT, Y INT -VE

2 OR BOTH Y, X + VE

FROM STATEMENT 1ST
AND FROM 2ND

WE GET TO KNOW Y INT IS -VE IN BOTH CASES

SO X INT IN BOTH CASES ARE -VE

D ANS