Cartesian Problem

[knight247]

Does line y=mx+r pass through the third and the fourth quadrant?
(1)m+r>0
(2)mr>0

[phoenix111]

Does line y=mx+r pass through the third and the fourth quadrant?
(1)m+r>0
(2)mr>0

It can be solved using statement (2) mr > 0 alone but not (1) alone.

For a line to pass through 3rd and 4th Quadrant the line must intersect the -ive X-axis
i.e. for y = 0 , x must have a unique -ive value.

from Eqn y = mx + r , and y = 0 , we get x = -r/m

since from stmt 2 : mr > 0 implies m & r have same sign,
so x = -r/m will always give x < 0 for mr>0.

[amit2k9]

a m+r > 0 meaning - y= 3x-2 | 3x+2 | -2x+3. in one case it passes and in another it doesn't.
hence insufficient.

b mr > < 0 meaning y= 3x+2 | -3x-2 . here too it passes in second case but not in first.
insufficient.

a+b only mr > 0 possible. meaning y= 3x+2. here it does not pass. Hence sufficient.

C it is.

[Geva@EconomistGMAT]

Does line y=mx+r pass through the third and the fourth quadrant?
(1)m+r>0
(2)mr>0

A line will pass through the 3rd and 4th quadrant if it hits the y axis below zero - meaning if r is negative. In this case, the slope does not matter - either positive, negative or zero, a line will pass from 3rd quadrant to 4th quadrant.


If r is positive and the line hits the y axis above the zero point, the line will miss one of the quadrants: if the slope is positive, the line will move from 3rd quad to 1st quad and bypass 4th, and if the slope is negative the line will move from 2nd to 4th and bypass 3rd. If the slope is zero, the line will miss both.

So this is the crux of the question: if r is negative, the answer is yes, but if r is positive, the answer is no.

Stat. (1) allows an r that is positive or negative, depending on the value of m. We could have both positive, or r negative and m positive to bring the sum back over zero. Insufficient.

stat (2) the same. We learn that m and r have the same sign - either both positive, or both negative. but we don't know whether r is positive or negative. Insufficient.

The combination is sufficient: stat. (2) says that m and r have the same sign - either both positive, or both negative. But if both negative, m+r cannot be >0 (stat. (1). Therefore, we get that both are positive, meaning that our line has a positive slope and intersects the y axis above zero - so it will miss out on quad 4. The answer to the question stem is NO, but that is sufficient. The answer is C.

[sl750]

Statement 1

m+r>0

If m is positive and r is positive. The line passes through 3rd quadrant
If m is positive and r is negative but the magnitude of r is smaller than m. The line passes through both 3rd and 4th quadrant. Insufficient

Statement 2

mr>0

Both m and r are positive. The line passes through the 3rd quadrant
Both m and r are negative. The lines passes through both 3rd and 4th quadrant. Insufficient

1+2

Both m and r are positive. The line passes through the 3rd quadrant. Sufficient