Problem Solving

Which quadrants, if any, contain points on the line y = x/1000+ 1,000,000?

This is a Manhattan problem. The answer is 1,2,3

The normal way of going about this is put x=0 and u get y=10^6 then put y=0 and get x=-10^9

When u draw the points (0,10^6) and (-10^9,0),u see that the line joining them passes thru quadrant 1,2 and 3. So great.

But here is where i have a problem. I've come across two thumb rules (1) if slope is -ve then the line passes thru quadrants 1,2 and 4 and (2) if slope is positive then the line passes thru quadrants 1,3 and 4

If I rearrange the terms of y = x/1000+ 1,000,000 then I get -x+1000y=10^9. Comparing this eqn with ax+by+c, I get, a=-1 and b=1000. Slope=-a/b=-(-1)/1000=1/1000 which is positive. So as per rule (2) shouldn't this line pass thru quadrants 1,3 and 4? Opinions?

I've come across two thumb rules (1) if slope is -ve then the line passes thru quadrants 1,2 and 4 and (2) if slope is positive then the line passes thru quadrants 1,3 and 4

Hi,
You must have missed other constraints. When the slope is negative, it will definitely pass through Q2 and Q4. But, whether it passes through Q1 or Q3 depends on the constant, c(y-intercept) in y=mx+c.
If c>0, it passes through quadrants 1,2,4.
If c<0, it passes through quadrants 2,3,4.
If c=0, it passes through Q2 and Q4 only.

Similarly, you can check for positive slope.

What Frankenstein wrote is absolutely right, I want to add to it the reasoning, since I think it is better to learn reasoning than to memorize rules.

A line with positive slope moves up as it moves to the right, and it moves down as it moves to the left. This means that eventually, no matter how long it takes, a line with positive slope will move up-right enough to be in quadrant 1, and down-left enough to be in quadrant 3.

Imagine that this positively-sloped line goes through the origin. In this case, it only moves through Quadrants 1 and 3. (Actually visualize this, don't just read the words.)

If we shift it up just a tiny bit, it will also go through quadrant 2, but never through 4.
Conversely, if we shift it down just a tiny bit, it will also go through quadrant 4, but never through 2.

The reverse of all of this is true with a negatively-sloped line:

As it moves to the right it moves down, and it moves up as it moves to the left. This means that eventually, no matter how long it takes, a line with negative slope will move down-right enough to be in quadrant 4, and up-left enough to be in quadrant 2.

Imagine that this positively-sloped line goes through the origin. In this case, it only moves through Quadrants 2 and 4. (Actually visualize this, don't just read the words.)

If we shift it up just a tiny bit, it will also go through quadrant 1, but never through 3.
Conversely, if we shift it down just a tiny bit, it will also go through quadrant 3, but never through 1.

Hope this is helpful.

Put x = 0
then Y = 0

knight247 wrote:Which quadrants, if any, contain points on the line y = x/1000+ 1,000,000?

This is a Manhattan problem. The answer is 1,2,3

The normal way of going about this is put x=0 and u get y=10^6 then put y=0 and get x=-10^9

When u draw the points (0,10^6) and (-10^9,0),u see that the line joining them passes thru quadrant 1,2 and 3. So great.

But here is where i have a problem. I've come across two thumb rules (1) if slope is -ve then the line passes thru quadrants 1,2 and 4 and (2) if slope is positive then the line passes thru quadrants 1,3 and 4

If I rearrange the terms of y = x/1000+ 1,000,000 then I get -x+1000y=10^9. Comparing this eqn with ax+by+c, I get, a=-1 and b=1000. Slope=-a/b=-(-1)/1000=1/1000 which is positive. So as per rule (2) shouldn't this line pass thru quadrants 1,3 and 4? Opinions?

Hi-- I understand that remembering theorems and rules is extremely important, but do we need to do all this in this question?

Looking at the question: isn't it obvious that if the value of X is Positive then Y can never be NEGATIVE ?

Hence, it can never belong to the 4th quadrant.

Yes, you can certainly look at the equation and confirm that (pos, neg) is never a possible value for (x, y). You should also double-check that each of the other three are possible (which they are).

I don't think that what I described actually takes all that long once you understand the trajectory of lines with positive slope and lines with negative slope.

But it's very helpful to think through your approach as well. This same approach is also helpful on DS questions some times. For example, if you see a - b > 0 you could conclude that it's not possible for both a < 0 and b > 0 to be true at the same time.

gmatboost wrote:Yes, you can certainly look at the equation and confirm that (pos, neg) is never a possible value for (x, y). You should also double-check that each of the other three are possible (which they are).

I don't think that what I described actually takes all that long once you understand the trajectory of lines with positive slope and lines with negative slope.

But it's very helpful to think through your approach as well. This same approach is also helpful on DS questions some times. For example, if you see a - b > 0 you could conclude that it's not possible for both a < 0 and b > 0 to be true at the same time.

yes, I totally agree with you.. actually I wanted knight247 to see this alternate approach as well...

thanks for method you described.. it helps