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

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
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VeritasKarishma Bunuel

I did exactly what Romen has done here.
What are we both missing?

Thanks for your time

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

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

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.

(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


(2): |x + y| = 1


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.

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.

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.

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.

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'.

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.

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.

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

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