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In the xy-coordinate plane, is point R equidistant from

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In the xy-coordinate plane, is point R equidistant from [#permalink]

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In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3) ?

(1) The x-coordinate of point R is -1.
(2) Point R lies on the line y = -3.
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In the xy-coordinate plane, is point R equidistant from [#permalink]

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In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3) ?

Look at the diagram below:

Image

Notice, that the green line (x=-1) is the perpendicular bisector of the line segment with endpoints (-3,-3) and (1,-3), thus ANY point on this line will be equidistant from points (-3,-3) and (1,-3).

(1) The x-coordinate of point R is -1 --> point R is on the green line. Sufficient.
(2) Point R lies on the line y = -3 --> point R may or may not be on the green line. Not sufficient.

Answer: A.

[Reveal] Spoiler:
Attachment:
Equidistant points.png
Equidistant points.png [ 9.68 KiB | Viewed 13656 times ]

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Re: In the xy-coordinate plane, is point R equidistant from [#permalink]

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Walkabout wrote:
In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3) ?

(1) The x-coordinate of point R is -1.
(2) Point R lies on the line y = -3.


Any point that lie on the perpendicular bisector the line segment with extreme points (-3,-3) and (1,-3) will satisfy this condition. The perpendicular bisector of the line segment is x=-1.

1) this means the point lies on x=-1. Sufficient.
2) This may or may not lie in the middle. The point -1,-3 is the mid point of the line segment but their are other points on the line such as (-2,-3) which doesn't satisfy the requirements. Insufficient.

Hence A.

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Re: In the xy-coordinate plane, is point R equidistant from [#permalink]

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New post 06 Aug 2014, 05:30
Why does the green line go through point -1 ? I couldn't find any chapter which explains this system.. can anyone help?

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In the xy-coordinate plane, is point R equidistant from [#permalink]

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We have to find out whether \(R(x,y)\) is equidistant from the two points mentioned

Using the distance formula
\((x+3)^2+(y+3)^2 = (x-1)^2+(y+3)^2\)
\((x+3)^2 = (x-1)^2\)

So basically we have to prove whether \((x+3)^2 = (x-1)^2\)or not?

1)Substituting\(-1\) in the above equation \((x+3)^2 = (x-1)^2\) results in it being equal
Thus sufficient

2)\(y = -3\)wouldn't help us with this eqn:\((x+3)^2 = (x-1)^2\)
Thus insufficient

Ans is A

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Re: In the xy-coordinate plane, is point R equidistant from [#permalink]

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New post 16 May 2015, 09:25
I got really scared seeing this question. But visualizing the coordinate plane made for a much simpler approach.

Once I got the two points mapped out it was obvious that point R had to be on X = -1.

I suppose this was possible because the two points had the same Y coordinates, which allowed for several a straight line at equidistance from the two points. Has anyone got any suggestions or Q's that involves points without this possibility? e.g. A=(1,0) B=(6,6)
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Re: In the xy-coordinate plane, is point R equidistant from [#permalink]

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Hi MarkusKarl,

Most Geometry questions have a "visual" component to them, so drawing the "work" involved (the shapes, the graph, etc.) will almost always be beneficial - in that way, you can connect conceptual ideas to real-world examples. GMAT questions in general are almost all pattern-based, so if you find yourself 'stuck' conceptually, you have to think about the rules involved and simplify the logic.

In your example, you name two points that don't share an X or Y co-ordinate, but the concept involved in this prompt applies to your example as well. There WILL be a "line" of co-ordinates that are equidistant from the two points that you named (it's just that the "line" will be a diagonal line and will NOT involve any shared X or Y co-ordinates).

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Re: In the xy-coordinate plane, is point R equidistant from [#permalink]

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Re: In the xy-coordinate plane, is point R equidistant from [#permalink]

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New post 04 Feb 2017, 00:45
Bunuel wrote:
Kchaudhary wrote:
In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3)?

(1) The x-coordinate of point R is -1.
(2) Point R lies on the line y=-3.


Merging topics. Please refer to the discussion above.


Hi Bunnel, in statement 1, how can you consider that y-coordinate of R is 0?

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Re: In the xy-coordinate plane, is point R equidistant from [#permalink]

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New post 04 Feb 2017, 00:55
Kchaudhary wrote:
Bunuel wrote:
Kchaudhary wrote:
In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3)?

(1) The x-coordinate of point R is -1.
(2) Point R lies on the line y=-3.


Merging topics. Please refer to the discussion above.


Hi Bunnel, in statement 1, how can you consider that y-coordinate of R is 0?


The y-coordinate of R is not necessarily 0. The point is that since x-coordinate of R is -1, then R is on the green line, so no matter what is the y-coordinate, R will be equidistant from the given points.

Image
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Re: In the xy-coordinate plane, is point R equidistant from [#permalink]

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New post 14 May 2017, 05:10
Hi lou34,

Thats because S1 says so.

S1 says - The X-co-ordinate of Point R is -1. This mean it will have to R WILL have to be ANY point on the green line mentioned by Bunuel.

I hope this clears your question.

lou34 wrote:
Why does the green line go through point -1 ? I couldn't find any chapter which explains this system.. can anyone help?

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Re: In the xy-coordinate plane, is point R equidistant from [#permalink]

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New post 14 May 2017, 05:55
HI Kchaudhary,

Just adding to Bunuel's solution.

Even though the term perpendicular bisector is used, we can infer S1 is sufficient without knowing about it. A quick sketch of the Co-ordinate will be able to get us there.

Consider the attached image. In the Image I have taken an Arbitrary point R on the green line. Named the given two points as A and B. Also labelled O for easier understanding.

What is asked is - AR = BR?. If you notice - Point A, B and R form two Right angles at common point O. In these two Right angled triangle, two sides are equal. AO = AB = 2 (Follows from given co-ordinates) and OR is common to both Right triangles. So, it clearly follows that the third side of both right angles MUST be equal. Meaning AR = BR. This makes S1 sufficient.

I hope this helped :)


Kchaudhary wrote:
Bunuel wrote:
Kchaudhary wrote:
In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3)?

(1) The x-coordinate of point R is -1.
(2) Point R lies on the line y=-3.


Merging topics. Please refer to the discussion above.


Hi Bunnel, in statement 1, how can you consider that y-coordinate of R is 0?

Attachments

File comment: Named given point A and B for clarity. Assumed Point R on green line
Equidistant points.png
Equidistant points.png [ 18.45 KiB | Viewed 2782 times ]


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In the xy-coordinate plane, is point R equidistant from [#permalink]

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New post 14 May 2017, 10:43
Walkabout wrote:
In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3) ?

(1) The x-coordinate of point R is -1.
(2) Point R lies on the line y = -3.


Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

From the original condition, we can set up as this question as follows.
The distance between \(R(x,y)\) and \((-3,-3)\) is \(\sqrt{(x+3)^2 + (y+3)^2}\) and the distance between \(R(x,y)\) and \((1,-3)\) is \(\sqrt{(x-1)^2 + (y+3)^2}\).
Thus, we have \(\sqrt{(x+3)^2 + (y+3)^2} = \sqrt{(x-1)^2 + (y+3)^2}\).
Then \(x^2 + 6x + 9 + y^2 + 6y + 9 = x^2 -2x + 1 + y^2 + 6y + 9\).
\(6x + 9 = -2x + 1\)
\(8x = -8\)
\(x = -1\)

We have 2 variables \(x\) and \(y\) and 1 equation, \(x = -1\).
In the original condition, there is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.

For 1), \(x = -1\), which is equivalent to the condition from the original question. No additional condition is provided. Thus this is not sufficient.

For 2), \(y = -3\). Then the point R is (-1,-3). This is sufficient.

Therefore, the answer is B.

-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: In the xy-coordinate plane, is point R equidistant from [#permalink]

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New post 10 Jul 2017, 08:04
MathRevolution wrote:
Walkabout wrote:
In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3) ?

(1) The x-coordinate of point R is -1.
(2) Point R lies on the line y = -3.


Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

From the original condition, we can set up as this question as follows.
The distance between \(R(x,y)\) and \((-3,-3)\) is \(\sqrt{(x+3)^2 + (y+3)^2}\) and the distance between \(R(x,y)\) and \((1,-3)\) is \(\sqrt{(x-1)^2 + (y+3)^2}\).
Thus, we have \(\sqrt{(x+3)^2 + (y+3)^2} = \sqrt{(x-1)^2 + (y+3)^2}\).
Then \(x^2 + 6x + 9 + y^2 + 6y + 9 = x^2 -2x + 1 + y^2 + 6y + 9\).
\(6x + 9 = -2x + 1\)
\(8x = -8\)
Got...methodical way.
\(x = -1\)

We have 2 variables \(x\) and \(y\) and 1 equation, \(x = -1\).
In the original condition, there is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.

For 1), \(x = -1\), which is equivalent to the condition from the original question. No additional condition is provided. Thus this is not sufficient.

For 2), \(y = -3\). Then the point R is (-1,-3). This is sufficient.

Therefore, the answer is B.

-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
Got it..methodical way.

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Re: In the xy-coordinate plane, is point R equidistant from [#permalink]

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New post 08 Aug 2017, 09:10
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Walkabout wrote:
In the xy-coordinate plane, is point R equidistant from points (-3,-3) and (1,-3) ?

(1) The x-coordinate of point R is -1.
(2) Point R lies on the line y = -3.


Target question: Is point R equidistant from points (-3,-3) and (1,-3)?
This question is a great candidate for rephrasing the target question.

First sketch the two given points
Image

Notice that the point (-1, -3) is equidistant from the two given points. MORE IMPORTANTLY, every point on the line x = -1 is equidistant from the two given points.
Image

So, we can rephrase the target question . . .
REPHRASED target question: Is point R on the line x = -1?

Statement 1: The x coordinate of point R is -1
If the x-coordinate is -1, then point R is definitely on the line x = -1
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: Point R lies on the line y= -3
This tells us nothing about whether or not point R is on the line x = -1?
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Answer:
[Reveal] Spoiler:
A


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Re: In the xy-coordinate plane, is point R equidistant from   [#permalink] 08 Aug 2017, 09:10
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