In the standard (x,y) coordinate plane, what is the slope of the line

Question

In the standard (x,y) coordinate plane, what is the slope of the line containing the distinct points P and Q ?

(1) Both P and Q lie on the graph of |x| + |y| = 1.
(2) Both P and Q lie on the graph of |x + y| = 1.

DS16291.01

nick1816 · Accepted Answer

Statement 1 - |x|+|y|=1
(1,0),(0,1), (-1,0) and (0,-1) satisfy equation

Case 1- 
If co-ordinates of P (-1,0) and that of Q is (0,1)
Slope=  \frac{1-0}{0-(-1)} = 1

Case 2- 
If co-ordinates of P (-1,0) and that of Q is (1,0)
Slope=  \frac{0-0}{1-(-1)} = 0

Insufficient

Statement 2 - |x+y|=1
Again (1,0),(0,1), (-1,0) and (0,-1) satisfy the equation

We can make the same cases

Case 1- 
If co-ordinates of P (-1,0) and that of Q is (0,1)
Slope=  \frac{1-0}{0-(-1)} = 1

Case 2- 
If co-ordinates of P (-1,0) and that of Q is (1,0)
Slope=  \frac{0-0}{1-(-1)} = 0

Insufficient

Combining both equations gives us nothing new, as those 4 points satisfies both equations.

Insufficient

E

KarishmaB · Answer

Look at the second diagram. P and Q could lie on the same line or on the two different lines. We need the slope of the line that joins P and Q. If the points lie on different lines, the slope of the line joining them could take any value.

romen2017 · Answer

Can it be B? 1. Graph for |x|+|y| =1 can have multiple slopes. NS. 2. Graph for |x+y|=1 would have single slope. Sufficient https://gmatclub.com/forum/download/fil ... 9848d3bf81

Karmesh · Answer

VeritasKarishma Bunuel I did exactly what Romen has done here. What are we both missing? Thanks for your time

BrentGMATPrepNow · Answer

Target question :    What is the slope of the line containing the distinct points P and Q ?

Statement 1 : Both P and Q lie on the graph of |x| + |y| = 1  
Since most people aren't familiar with the graph of |x| + |y| = 1, it makes sense to  test some values 
There are infinitely many values of x and y that satisfy statement 1. Here are two: 
 Case a : For point P, x = 1 and y = 0 (1,0) and, for point Q, x = 0 and y = 1 (0,1). In this case,  the slope of the line = (1 - 0)/(0 - 1) = -1 
 Case b : For point P, x = -1 and y = 0 (-1,0) and, for point Q, x = 0 and y = 1 (0,1). In this case,  the slope of the line = (1 - 0)/(0 - -1) = 1 
Since we cannot answer the  target question  with certainty, statement 1 is NOT SUFFICIENT

Statement 2 : Both P and Q lie on the graph of |x + y| = 1 
Before we do anything else, let's first check to see whether we can  re-test  the same values we used to show that statement 1 is not sufficient. 
Yes, it turns out that we CAN re-use those same values to get: 
 Case a : For point P, x = 1 and y = 0 (1,0) and, for point Q, x = 0 and y = 1 (0,1). In this case,  the slope of the line = (1 - 0)/(0 - 1) = -1 
 Case b : For point P, x = -1 and y = 0 (-1,0) and, for point Q, x = 0 and y = 1 (0,1). In this case,  the slope of the line = (1 - 0)/(0 - -1) = 1 
Since we cannot answer the  target question  with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined    
Since we were able to use the  same counter-examples  to show that each statement ALONE is not sufficient, the same counter-examples will satisfy the two statements COMBINED. 
In other words, 
 Case a : For point P, x = 1 and y = 0 (1,0) and, for point Q, x = 0 and y = 1 (0,1). In this case,  the slope of the line = (1 - 0)/(0 - 1) = -1 
 Case b : For point P, x = -1 and y = 0 (-1,0) and, for point Q, x = 0 and y = 1 (0,1). In this case,  the slope of the line = (1 - 0)/(0 - -1) = 1 
Since we cannot answer the  target question  with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent

suminha · Answer

Hi, Brent, thank you for the precise explanation.

I have a question, 
I think I’ve seen somewhere that when we see an equation such as this,  |x|+|y|=1 , we can simply calculate in this way:  x+y=1 , because anyway those are positives. 
So I thought that I can make the equation  y=-x+1 . Thus the slope is  -1 .

What is wrong with thinking this way and do you know in what kind of situation we should consider  |x|+|y|=1   x+y=1 ?

Thank you in advance!

Posted from my mobile device

BrentGMATPrepNow · Answer

We can quickly see that  |x|+|y|=1  and  x+y=1  are not equivalent equations.
For example, the values x = -1 and y = 0 satisfy the first equation but not the second.

One instance in which we can replace  |x|+|y|=1  with  x+y=1  would be if we're told that x and y are both positive. I'm not sure about other instances.

Bambi2021 · Answer

(1) will only give us one type of graph, a quadrilateral connected by the points (0,1), (0,-1), (1,0) and (-1,0)

(2) is a little more complex. On one hand we get the same quadrilateral as in (1). But on the other we also get one upward sloping line that connects the dots (-0,5,-0,5) and (0,5, 0,5), and one downward sloping line that connects the dots (-0,5, 1,5) and (0,5,-1,5).

Edit: Actually, Im not sure whether (2) gives us any extra lines at all. They must in any case be broken for values less than 1. Am I correct?

Edit 2: I get the following graphs. It seems like (2) should be sufficient? What is not shown on my graphs?

(1): |x| + |y| = 1
 https://i.imgur.com/d1qli0K.jpg

(2): |x + y| = 1
 https://i.imgur.com/nrN5ncs.jpeg

Edit 3: 
Of course! Now I get it. We only know that both P and Q lie on the graph, but we dont know on which of the lines.

CrackverbalGMAT · Answer

In the standard (x,y) coordinate plane, what is the slope of the line containing the distinct points P and Q?

(1) Both P and Q lie on the graph of |x| + |y| = 1.
(2) Both P and Q lie on the graph of |x + y| = 1.

Really nice DS question that tests your understanding of slope and absolute values as well as your logical reasoning skills.

Let's start by understanding the Question stem. P and Q are distinct points and we need to find the slope of the line containing P and Q. So it's a  unique value-based DS question .

(1) Both P and Q lie on the graph of  |x| + |y| = 1.

Do you think the graph |x| + |y| = 1 will represent only a single straight line?   No .

Definitely, the graph will have more than 1 line due to the properties of absolute values. |x| can be x or -x depending on the sign of x . The same can be applied to |y|.
 Logically thinking, this info is enough to conclude that Statement 1 is not sufficient.

Because there are 2 possible cases for points P and Q.

Case 1 : P and Q are distinct points in the same line 1.

Case 2:  P and Q are in Line 1 and line 2 respectively and the slope of the line connecting P and Q will be different from the slope in case 1.

Since we are not able to find a unique value for slope ,  Statement 1 alone is not sufficient .

(2) Both P and Q lie on the graph of  |x + y| = 1.
 
The graph of |x + y| = 1 will also have two straight lines. Based on the definition of modulus, the graph will have 2 lines.
Line 1:  x + y = 1  if x+y ≥ 0
Line 2 : -(x+ y) = 1 ==>  -x-y = 1   if x+y < 0.

Due to the same reasons mentioned in statement 1, a unique value for the slope is not possible. Hence,   statement 2 alone is also insufficient.

The next option we have is to combine both statements. Even though if you combine, we should be able to come up with a single common line segment in order to get a unique value for slope.

Let's analyze St 1. in detail to proceed further. |x| + |y| = 1.
|x| and |y| will behave differently based on each quadrant.

Case 1: x ≥ 0 and y ≥ 0 :  Quadrant 1 
  x + y = 1

Case 2: x < 0 and y ≥ 0 : Quadrant 2

-x + y = 1

Case 3: x < 0 and y < 0 :  Quadrant 3

-x - y = 1

Case 4: x ≥ 0 and y < 0: Quadrant 4
x - y = 1

So, the graph will contain a different line segments in each quadrant .So a total of 4 line segments are there. You can also try to plot the graph using x and y-intercepts of each line for better understanding.

Combining both statements, we can see that 2 line segments are common in both graphs (St 1 and St.2) in quadrants 1 and 3.
x + y = 1 in quadrant 1 and -x - y = 1 in quadrant 3.

So, the points P and Q can lie in the same line segment or in the different ones. 
Due to the same reasons mentioned earlier, since we have 2 different line segments, getting a unique value for the slope of the line connecting P and Q is not possible. Therefore,  option E is the correct answer.

Thanks,
Clifin J Francis,
GMAT Mentor.

ranforce · Answer

hey can any one help me out with statement two?

I calculated Ix+yI =1; as

case a : x+y =1, therefore, y= -x+1 slope here is -1

case b x+y = -1, therefore, y= -x -1 slope here is -1.

Where I am wrong in my reasoning?

if there is any fundamental concept I missed please advise on same. I will appreciate links to concept or anything this may help me understand.

Best.

godhand · Answer

I followed an algebraic/inequality based approach to get my answer. Can anyone know if I am right?

1) and 2) alone are insufficient due to reasons others have stated.

When we take 1 & 2 together, we get:

|x| + |y| = 1 = |x+y|

i.e |x| + |y| = |x+y|

According to inequality rules, |x+y| is always <= |x| + |y|.

BUT |x| + |y| = |x+y| if one of the below two criteria are met:

a) either x or y is zero, or both are zero (coming back to the question - if this is the case, slope is zero)
OR
b) both x and y have the same sign (in this case, a slope can be determined. Say if x=5 and y=7, or x=-5 and y=-7)

Since we have conflicting answers, the solution is E.

dushyantkanal · Answer

Hi

I got E as the answer, however, just wanted to check if the method used is correct.

PFA, my working of the two statements.

St 1 - 4 lines are possible, 2 different slopes, P and Q can be anywhere - Hence Not Sufficient
St 2 - 2 lines are possible, 2 different slopes, P and q can be anywhere - Hence Not Sufficient

Together, we get the St 2 lines as common, which are known to be insufficient.

Hence answer is 'E'.

tabish314 · Answer

I did the same thing and in statement 2) we get two equations, namely
y = -x + 1
y = -x -1

Q asks the slope of the line which contains PQ. 
How did you arrive to the conclusion that statement 2) has multiple slopes as both equations show slope to be -1 although the y-intercepts are different (1 and -1) ?
Also to me it seems that both the lines you drew for statement 2) are parallel lines and parallel lines have same slopes.

happypuppy · Answer

Just that in the 2nd graph, P can be on one line and Q can be on another line. 
Hence can have a diff slope than the graph's slope.

gag1994 · Answer

SaquibGMATWhiz  How is B not sufficient?

Through B we get 2 equations with x+y=1 and x+y=-1, meaning the slope is always -1.

Regards

Gagan

hr1212 · Answer

GMAT-Club-Forum-3bah7kvc.png As multiple points can be selected on the intersection of both the graphs (where red and blue overlaps) with different slopes both statements together are not sufficient. IMO: E

In the standard (x,y) coordinate plane, what is the slope of the line

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

In the standard (x,y) coordinate plane, what is the slope of the line

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards