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y=3x+2 - point (r,s)

Problem Solving — algebra and arithmetic
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by vishal_2804 » Wed Apr 24, 2013 3:25 am
In the xy plane, does the line with equation y=3x+2 contain the point (r,s)?

1) (3r+2-s)(4r+9-s)=0
2) (4r-6-s)(3r+2-s)=0.

Pls explain the step and the logic to solve similar type of problems.
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by Anju@Gurome » Wed Apr 24, 2013 3:54 am
vishal_2804 wrote:In the xy plane, does the line with equation y=3x+2 contain the point (r,s)?

1) (3r+2-s)(4r+9-s)=0
2) (4r-6-s)(3r+2-s)=0
If the point (r, s) is on the line y = 3x + 2, then r = 3r + 2 ---> (3r + 2 - s) = 0
So the question reduces to "Is (3r + 2 - s) = 0?"

Statement 1: (3r + 2 - s)(4r + 9 - s) = 0
So, either (3r + 2 - s) = 0 OR (4r + 9 - s) = 0

Not sufficient


Statement 2: (4r - 6 - s)(3r + 2 - s) = 0
So, either (4r - 6 - s) = 0 OR (3r + 2 - s) = 0

Not sufficient

1 & 2 Together: Either {(4r + 9 - s) = 0 and (4r - 6 - s) = 0} OR (3r + 2 - s) = 0
Now, if (4r + 9 - s) = 0 and (4r - 6 - s) = 0
--> (4r - s) = -9 and (4r - s) = 6
--> This is not possible for any values of r and s

So, (3r + 2 - s) must be equal to zero.

Sufficient

The correct answer is C.
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by Brent@GMATPrepNow » Wed Apr 24, 2013 6:38 am
vishal_2804 wrote:In the xy plane, does the line with equation y=3x+2 contain the point (r,s)?

1) (3r+2-s)(4r+9-s)=0
2) (4r-6-s)(3r+2-s)=0.
If (r,s) is on the line defined by the equation y=3x+2, then (r,s) must satisfy the equation y=3x+2. In other words, it must be true that s=3r+2
For example: We know that the point (5, 17) is on the line y=3x+2, because when we plug x=5 and y=17 into the equation, we get 17 = 3(5)+2 and the equation holds true.

So, we can reword the target question to be "Does s = 3r + 2?"

1. (3r+2-s)(4r+9-s) = 0
From this, we know that either (3r+2-s) = 0 or (4r+9-s) = 0
If (3r+2-s) = 0 then s = 3r+2, in which case the answer to our new target question is yes
If (4r+9-s) = 0 then s = 4r+9, in which case the answer to our new target question is no
Since we get two different answers to the target question, statement 1 is NOT SUFFICIENT

2. (4r-6-s)(3r+2-s) = 0
From this, we know that either (4r-6-s) = 0 or (3r+2-s) = 0
If (4r-6-s)) = 0 then s = 4r-6, in which case the answer to our new target question is no
If (3r+2-s) = 0 then s = 3r+2, in which case the answer to our new target question is yes
Since we get two different answers to the target question, statement 2 is NOT SUFFICIENT

Statements 1&2 combined: Since (3r+2-s) is the only expression common to both statements, it must be true that 3r+2-s = 0, in which case s MUST equal 3r+2
As such the answer is C

Cheers,
Brent
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by srcc25anu » Tue Apr 30, 2013 10:48 am
Question is whether point (r,s) lies on RED line ?

St1: Either point (r,s) lies on RED line or GREEN line.
Not sufficient

St2: Either point (r,s) lies on RED line or BLUE line.
Not sufficient

Together: Only possible scenario is that the point lies on red line. No one point can possibly lie on both the green and blue line because they are parallel. They will never intersect each other.

Hence Point (r,s) lies on RED line.
SUFFICIENT

Ans C
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by jervizeloy » Sun Oct 30, 2016 6:39 am
Brent@GMATPrepNow wrote:
vishal_2804 wrote:In the xy plane, does the line with equation y=3x+2 contain the point (r,s)?

1) (3r+2-s)(4r+9-s)=0
2) (4r-6-s)(3r+2-s)=0.
If (r,s) is on the line defined by the equation y=3x+2, then (r,s) must satisfy the equation y=3x+2. In other words, it must be true that s=3r+2
For example: We know that the point (5, 17) is on the line y=3x+2, because when we plug x=5 and y=17 into the equation, we get 17 = 3(5)+2 and the equation holds true.

So, we can reword the target question to be "Does s = 3r + 2?"

1. (3r+2-s)(4r+9-s) = 0
From this, we know that either (3r+2-s) = 0 or (4r+9-s) = 0
If (3r+2-s) = 0 then s = 3r+2, in which case the answer to our new target question is yes
If (4r+9-s) = 0 then s = 4r+9, in which case the answer to our new target question is no
Since we get two different answers to the target question, statement 1 is NOT SUFFICIENT

2. (4r-6-s)(3r+2-s) = 0
From this, we know that either (4r-6-s) = 0 or (3r+2-s) = 0
If (4r-6-s)) = 0 then s = 4r-6, in which case the answer to our new target question is no
If (3r+2-s) = 0 then s = 3r+2, in which case the answer to our new target question is yes
Since we get two different answers to the target question, statement 2 is NOT SUFFICIENT

Statements 1&2 combined: Since (3r+2-s) is the only expression common to both statements, it must be true that 3r+2-s = 0, in which case s MUST equal 3r+2
As such the answer is C

Cheers,
Brent
Hi Brent,

Just a minor question that does not change the result but would be of great help to reinforce my concepts.

I think for each statement there are three cases to analyze. For example, for statement 1 we would have:

3r+2-s=0

OR

4r+9-s=0

OR

3r+2-s=0 AND 4r+9-s=0

Would such analysis be correct?
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by Brent@GMATPrepNow » Sun Oct 30, 2016 2:30 pm
jervizeloy wrote: Hi Brent,

Just a minor question that does not change the result but would be of great help to reinforce my concepts.

I think for each statement there are three cases to analyze. For example, for statement 1 we would have:

3r+2-s=0

OR

4r+9-s=0

OR

3r+2-s=0 AND 4r+9-s=0

Would such analysis be correct?
Yes, that's a correct analysis.

Cheers,
Brent
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by Matt@VeritasPrep » Fri Nov 11, 2016 1:38 pm
jervizeloy wrote: I think for each statement there are three cases to analyze. For example, for statement 1 we would have:

3r+2-s=0

OR

4r+9-s=0

OR

3r+2-s=0 AND 4r+9-s=0

Would such analysis be correct?
This is true, but the cases are different in an important way. The first two give you r in terms and s, and the last one gives you the exact values of r and s. So depending on the prompt, the first two might be sufficient, or you might need the third case.
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