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Re: In the XY plane, does the line with equation y=3x+2 [#permalink]
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30 Aug 2013, 15:41
Bunuel,
In statement (1), you have said that we have these possibilities: (3r + 2  s) = 0 OR (4r + 9  s) = 0 OR both.
But, when we have something like this: (x + 3)(x+5) = 0
In this case, there are only two possibilities, right? (x+3)=0 OR (x+5)=0. Both cannot be zero at the same time because "x" represents a single and unique value. Please confirm.



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Re: In the XY plane, does the line with equation y=3x+2 [#permalink]
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31 Aug 2013, 06:13



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Re: In the XY plane, does the line with equation y=3x+2 [#permalink]
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07 Jun 2014, 12:40
could we also think of it as:
the following two equations define two nonidentical hyperbola. 1) (3r+2s) (4r+9s)=0 2) (4r6s) (3r+2s)=0
each equation represents a set of points on different hyperbola since the question states that (r,s) is one point, it must be a single intersection of the two hyperbola which cannot be found without both equations.
therefore: you need both equations to define one point (r,s) and then you can easily solve for whether (r,s) satisfies the first equation y = 3x+2
would that reasoning also work?



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In the XY plane, does the line with equation y=3x+2 [#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+2s)(4r+9s)=0 and (4r6s)(3r+2s)=0, there are 2 slightly unrelated terms like 5r3s and 2r+7s.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.



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Re: In the XY plane, does the line with equation y=3x+2 [#permalink]
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26 Nov 2014, 05:46



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In the xy plane, does the line y=3x+2 contain the point (r,s)? i) (3r+ [#permalink]
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01 Aug 2015, 23:38
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In the xy plane, does the line y=3x+2 contain the point (r,s)? i) (3r+2s)(4r+9s)=0 ii) (4r6s)(3r+2s)=0



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Re: In the xy plane, does the line y=3x+2 contain the point (r,s)? i) (3r+ [#permalink]
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02 Aug 2015, 03:24
visram04 wrote: In the xy plane, does the line y=3x+2 contain the point (r,s)? i) (3r+2s)(4r+9s)=0 ii) (4r6s)(3r+2s)=0 Question Stem: If Point (r,s) lies on the line y =3x+2, then the point should satisfy the equation, i.e., s=3r+2St 1: (3r+2s)(4r+9s)=03r+2 = s or 4r+9 =s We dont have unique solution. Hence not sufficient St 2: (4r6s)(3r+2s)=04r6 = s or 3r+2 = s We dont have unique solution. Hence not sufficient Combined:Combining we get s = 3r+2 > which is required to prove from the question stem. Hence sufficient.
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In the XY plane, does the line with equation y=3x+2 [#permalink]
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22 Mar 2016, 04:32
gaurav2k101 wrote: In the xyplane , does the line with equation y=3x+2 contains the point(r,s)?
(1) (3r+2s)(4r+9s)=0 (2) (4r6s)(3r+2s)=0 I did it differently, but it could be totally wrong. It would be good if someone can check it. Bunuel I would highly appreciate it if you check it. I substituted 0 for r for both statements, (1) (3(0)+2s)(4(0)+9s)=0 so that s=2 (which is on the line) or 9 (which is not on the line) (2) (4(0)6s)(3(0)+2s)=0 so that s=6 (which is not on the line) or 2 (which is on the line) Alone statement 1 and 2 are insufficient since they could be either 1 of 2 values, but together they prove that s=2, so the answer is C.



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Re: In the XY plane, does the line with equation y=3x+2 [#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.



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Re: In the XY plane, does the line with equation y=3x+2 [#permalink]
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20 Aug 2016, 10:25
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Guys, please say if I am wrong. If the question was found that 1) (3r+3s) (4r+9s)=0 2) (4r6s) (3r+3s)=0 then we will conclud that the answer is C, do not contain since parallel
If the question was found that 1) (4r+2s) (4r+9s)=0 2) (4r6s) (4r+2s)=0 then we will conclud that the answer is C, contain since intercept
am I right?



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Re: In the XY plane, does the line with equation y=3x+2 [#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: 3rs+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. (4rs = 9 as well as 6> not possible) so only case 1 holds i.e. (3r + 2  s) is equal to zero (C) is the answer.
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Re: In the XY plane, does the line with equation y=3x+2 [#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 \(3r+2s=0\)) holds true.
So basically we are asked to determine whether \(3r+2s=0\) is true or not.
(1) \((3r+2s)(4r+9s)=0\) > either \(3r+2s=0\) OR \(4r+9s=0\) OR both. Not sufficient.
(2) \((4r6s)(3r+2s)=0\) > either \(3r+2s=0\) OR \(4r6s=0\) OR both. Not sufficient.
(1)+(2) Both \(4r+9s=0\) and \(4r6s=0\) cannot be true (simultaneously), as \(4rs\) can not equal to both 9 and 6, hence \(3r+2s=0\) 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+2s=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.



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Re: In the XY plane, does the line with equation y=3x+2 [#permalink]
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01 Feb 2017, 08:06
Kapidhwaj wrote: 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 \(3r+2s=0\)) holds true.
So basically we are asked to determine whether \(3r+2s=0\) is true or not.
(1) \((3r+2s)(4r+9s)=0\) > either \(3r+2s=0\) OR \(4r+9s=0\) OR both. Not sufficient.
(2) \((4r6s)(3r+2s)=0\) > either \(3r+2s=0\) OR \(4r6s=0\) OR both. Not sufficient.
(1)+(2) Both \(4r+9s=0\) and \(4r6s=0\) cannot be true (simultaneously), as \(4rs\) can not equal to both 9 and 6, hence \(3r+2s=0\) 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+2s=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+2s)(4r+9s)=0 and (4r6s)(3r+2s)=0 has is 3r+2s=0, which makes the answer C correct. Also, note that this is an official question and confirmed OA is C.
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