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Re: In the xy-plane , does the line with equation y = 3x + 2 contains the [#permalink]

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25 Nov 2014, 12:28

(Just so I get the whole picture) Say, instead of the bold parts in (3r+2-s)(4r+9-s)=0 and (4r-6-s)(3r+2-s)=0, there are 2 slightly unrelated terms like 5r-3-s and 2r+7-s.They would remain to be possibilities of the equation and E would be the right choice, right? Sorry if I am complicating things.

Last edited by deeuk on 26 Nov 2014, 11:36, edited 1 time in total.

(Just so I get the whole picture) Say, instead of the bold parts in (3r+2-s)(4r+9-s)=0 and (4r-6-s)(3r+2-s)=0, there were 2 slightly unrelated terms like 5r-3-s and 2r+7-s, they would remain to be possibilities of the equation and E would be the right choice, right? Sorry if I am complicating things.

Yes, in this case the answer would be E.
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Re: In the xy-plane , does the line with equation y = 3x + 2 contains the [#permalink]

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05 Aug 2016, 10:07

Substituting variables r and s for x and y is very smart. Then the question becomes a logic question.

(1) can contain the point or it could not (2) can contain the point or could not (just like statement 1)

Together at first it looks like it will not work out, however it has to have the point because the other two equations could not satisfy both of the equations. It would be 4r + 9 = 4r - 6 which would never work.

Re: In the xy-plane , does the line with equation y = 3x + 2 contains the [#permalink]

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20 Aug 2016, 10:25

1

This post was BOOKMARKED

Guys, please say if I am wrong. If the question was found that 1) (3r+3-s) (4r+9-s)=0 2) (4r-6-s) (3r+3-s)=0 then we will conclud that the answer is C, do not contain since parallel

If the question was found that 1) (4r+2-s) (4r+9-s)=0 2) (4r-6-s) (4r+2-s)=0 then we will conclud that the answer is C, contain since intercept

Re: In the xy-plane , does the line with equation y = 3x + 2 contains the [#permalink]

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30 Jan 2017, 18:24

(r,s) will lie on the line with equation y = 3x + 2, if in place of x and y respectively we put the coordinates of the point and the line equation is satisfied.

So , (r,s) will lie on y = 3x +2, if s=3r+2 or in other words:

3r-s+2 = 0.

From St. 1:

(3r + 2 - s)(4r + 9 - s) = 0 i.e.

either

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

Say (3r + 2 - s) = 0 -->St. 1 satisfied and (r, s) lies on given line.

Now say, (3r + 2 - s) not equal to zero, but (4r + 9 - s) = 0

Then St. 1 satisfied but (r, s) does not lie on given line.

So St. 1 is not sufficient.

Exactly same logic for St. 2.

St. 2 not sufficient.

Combining both statements.

case 1: (3r + 2 - s) is equal to zero

case 2: (3r + 2 - s) is not equal to zero

let's discuss case (2):

This means that

(4r + 9 - s) = 0 and also (4r - 6 - s) = 0

which is absurd. (4r-s = -9 as well as 6--> not possible)

Re: In the xy-plane , does the line with equation y = 3x + 2 contains the [#permalink]

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01 Feb 2017, 07:33

Bunuel wrote:

In the XY plane, does the line with equation y=3x+2 contain the point (r,s)?

Line with equation \(y=3x+2\) contains the point \((r,s)\) means that when substituting \(r\) ans \(s\) in line equation: \(s=3r+2\) (or \(3r+2-s=0\)) holds true.

So basically we are asked to determine whether \(3r+2-s=0\) is true or not.

(1) \((3r+2-s)(4r+9-s)=0\) --> either \(3r+2-s=0\) OR \(4r+9-s=0\) OR both. Not sufficient.

(2) \((4r-6-s)(3r+2-s)=0\) --> either \(3r+2-s=0\) OR \(4r-6-s=0\) OR both. Not sufficient.

(1)+(2) Both \(4r+9-s=0\) and \(4r-6-s=0\) cannot be true (simultaneously), as \(4r-s\) can not equal to both -9 and 6, hence \(3r+2-s=0\) must be true. Sufficient.

Answer: C.

Hi,

This solution will not hold true if r and s are on a rotated axis on the plane i.e. r= xcosA + ysinA and s = -xsinA + ycosA for any A. In this case line representing 3r+2-s=0 will intersect 3x+2 = y at a single point, thus the line 3x+2 = y may or may not contain (r,s), hence, E.

In the XY plane, does the line with equation y=3x+2 contain the point (r,s)?

Line with equation \(y=3x+2\) contains the point \((r,s)\) means that when substituting \(r\) ans \(s\) in line equation: \(s=3r+2\) (or \(3r+2-s=0\)) holds true.

So basically we are asked to determine whether \(3r+2-s=0\) is true or not.

(1) \((3r+2-s)(4r+9-s)=0\) --> either \(3r+2-s=0\) OR \(4r+9-s=0\) OR both. Not sufficient.

(2) \((4r-6-s)(3r+2-s)=0\) --> either \(3r+2-s=0\) OR \(4r-6-s=0\) OR both. Not sufficient.

(1)+(2) Both \(4r+9-s=0\) and \(4r-6-s=0\) cannot be true (simultaneously), as \(4r-s\) can not equal to both -9 and 6, hence \(3r+2-s=0\) must be true. Sufficient.

Answer: C.

Hi,

This solution will not hold true if r and s are on a rotated axis on the plane i.e. r= xcosA + ysinA and s = -xsinA + ycosA for any A. In this case line representing 3r+2-s=0 will intersect 3x+2 = y at a single point, thus the line 3x+2 = y may or may not contain (r,s), hence, E.

Dear Kapidhwaj,

Not sure what are you trying to say but the only solution (3r+2-s)(4r+9-s)=0 and (4r-6-s)(3r+2-s)=0 has is 3r+2-s=0, which makes the answer C correct. Also, note that this is an official question and confirmed OA is C.
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