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# ds:geometry

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10 Apr 2006, 05:56
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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
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10 Apr 2006, 07:22
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

Lets look at the question first. We have an equation giving relationship between x & y. And the question asks us if (r,s) is contained in y=3x+2

Usually when we want to see if if something is contained in a given line equation we have to sub those values and see if the equation holds good.
Ah so we should be looking for values of r & s in statements 1 and 2

Good now armed with this information lets look at statement 1
(3r+2-s)(4r + 9 -s) = 0
Great they already factored it for us so
3r+2-s = 0 --> 3r-s = -2 ------------Eqn(1)
4r+9-s = 0 ----> 4r-s = -9 ----------Eqn(2)

Solving Eqn (1) and Eqn(2) we get r= -7 and s= -19

Sub (r,s) in the question stem y=3x+2 ---> -19 = 3(-7) +2--> -19 = -19

Great Statement 1 is sufficient

Statement(2)
(4r -6 -s )(3r + 2 - s) = 0
This one is factored as well.
(4r -6 -s ) = 0 -------> 4r-s = 6 ------Eqn(3)
(3r+2-s) = 0 ------> 3r-s = -2 ------Eqn(4)

Solving Eqn (3) and Eqn (4) we get r = 8 , s= 26

Substitute in question stem
y= 3x + 2 --> 26 = 3(8) + 2 ---> 26 = 26

Statement 2 is sufficient as well.

Hence

Heman & the master of the universe

Last edited by heman on 16 Sep 2006, 17:21, edited 2 times in total.
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10 Apr 2006, 07:44
heman,

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10 Apr 2006, 07:45
macca 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

s= 3r +2

from 1
3r + 2 = s

s = 4r + 9

From these two statements we can deduce that (r,s) may or may not be on line

From 2

s = 4r - 6
s = 3r+2

Again (r,s) may or may not be on the line.

Combining 1 and 2
There is only one equaltion on which (r,s) lies

hence C
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10 Apr 2006, 07:50
You need both the Choices to arrive at one solution. Hence C.

- Vipin
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10 Apr 2006, 08:27
gmat_crack Can you tell me what is the flaw in my solution.?

Heman
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10 Apr 2006, 10:55
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

Q, y=3x+2, intersects y axis at 2 and slope of 3 (imagine)

s=3r+2 (same line equation, definetely intersects), may or may not intersect
s=4r+9 (intersects y axis at 9 and more slope) so it has to intersect line in Q

From 2,
s = 4r-6(intersects y axis on -6 and slope of 4), may or may not intersect
s=3r+2(same line equation) definetely intersects

Does this make sense? Am i thinking in right direction?

Both ways we cannot say. I think answer is E.

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10 Apr 2006, 12:05
Hope we can get some help from HongHu!
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10 Apr 2006, 22:21
My guess is B.
For the points (r,s) to lie on the line y=3x+2, the slope of r,s should be equal to the slope of y=3x+2 and since 2 provids us with this (s=3r+2) my guess is B.
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10 Apr 2006, 22:30

in the two equations substitute s = 3r+2

if the LHS == 0 then the equation holds true.

U can see its true for both 1 and 2.

Hence D
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11 Apr 2006, 00:53
pseudo wrote:
My guess is B.
For the points (r,s) to lie on the line y=3x+2, the slope of r,s should be equal to the slope of y=3x+2 and since 2 provids us with this (s=3r+2) my guess is B.

hmm, but the point (r,s) could be on a line parallell to y=3x+2?
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11 Apr 2006, 00:55
macca wrote:
pseudo wrote:
My guess is B.
For the points (r,s) to lie on the line y=3x+2, the slope of r,s should be equal to the slope of y=3x+2 and since 2 provids us with this (s=3r+2) my guess is B.

hmm, but the point (r,s) could be on a line parallell to y=3x+2?

I am not sure whether you can just use the slope!

You have to prove that there are/is atleast 1 point where they intersect!
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11 Apr 2006, 12:25
1, (3r+2-s)(4r + 9 -s) = 0

2, (4r -6 -s )(3r + 2 - s) = 0

It is C.

You cannot simply plug in s = 3r +2. You have to determine whether the equations given have a root that is common with s = 3r + 2.

Statement 1:

s = 4r + 9 OR 3r + 2. Therefore insufficient.

Statement 2:

s = 4r - 6 OR 3r + 2. Also insufficient.

Statements 1 and 2: Taking both equations, we know that s MUST equal 3r + 2, since it is a root to both equations. Therefore C.

If you simply substitute s = 3r + 2, of course you'll get both sides equal to zero, because it is one of the two roots of each equation. This does not necessarily mean that both statements are insufficient by themselves.
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11 Apr 2006, 17:01
jcgoodchild wrote:
You cannot simply plug in s = 3r +2. You have to determine whether the equations given have a root that is common with s = 3r + 2.

Well it depends. You can plug in s = 3r +2. In essense what you are doing is illustrating that at that point the same values of s,r makes sense for both the equation. Though, you cannot say that either equation 1 or 2 is equivalent to 0.

However, if it is equivalent to 0, you can show that the two equations are valid for the same values of s,r! I.e. the two curve/lines intersect.

Another way to explain the above would be to draw the graph of the two equations!

If they intersect, they will have the same values of s,r satisfying the two equations!
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12 Apr 2006, 01:50
macca 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

A) (3r+2-s)(4r + 9 -s) = 0
=> (3r+2-s) = 0 or (4r + 9 -s)=0
so any pair of (r,s) which satisfy one of above will solve Eq-A
In sufficient

B) Also in sufficient with same logic

A+B to gather means only 3r+2-s=0 (s=3r+2) is valid option Hence C is the answer.

-------------------------------------------------
Other way to visualize this problem is to consider r and s as variables, this means 3r+2-s=0 is a Line(line1), so does 4r+9-s=0(line2), so any value on line 1 or line 2 will satifsy eq. one so we really dont know which exact line we are talking about.

Same logic will apply to B

But if we see A, B simultaniously only points on Line 3r+2-s=0 will satisfy both equation, hence C is the answer.
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12 Apr 2006, 06:27
OA is B!

Don't understand how to solve this problem.
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12 Apr 2006, 08:00
macca wrote:
OA is B!

Don't understand how to solve this problem.

What is source of your question?
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12 Apr 2006, 08:23
The source is gmatprep
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12 Apr 2006, 13:22
The official answer is wrong and I say this full authority.

In order for a point with coordinates (r,s) to lie on a line y=3x+2, the slope between the point (r,s) and another point lying on line y-3x+2 should be 3.

Lets say x =1 then y= 5

So slope between (R,S) and (1,5) should be 3.

Now lets take B

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

Which means that 4r-s-s =0 and also 3r+2-s=0

Simply those so that

r= (s+6)/4 â€¦. (1)
s= s = 3r+2â€¦. (2)

now take the slope between (R,S) and (1,5)

slope = s-5 / r-1

substiture r and s from 1 and 2 and you will get:

slope = 12(r-1) / (s+2)

substitute s from 2

slope = 12(r-1) / 3r+4

you are stuck and you cannot find the slope. So B can not be the answer.
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12 Apr 2006, 21:00
Do they have any explanation? This sounds weird!
12 Apr 2006, 21:00

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# ds:geometry

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