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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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Re: 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:
Attachment:
Equidistant points.png
Equidistant points.png [ 9.68 KiB | Viewed 7609 times ]
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.
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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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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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New post 16 May 2015, 11:32
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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] 16 May 2015, 11:32
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