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

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

Look at the diagram below:



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.


_________________

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


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.


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 point R is on the line x = -1?
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

RELATED VIDEO

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.

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)

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).

GMAT assassins aren't born, they're made,
Rich

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.

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 line y = -3.

We get a graph as below:


In other words, it is asking whether point R is in the same distance from (-3,-3) and (1,-3).
The line x=-1 is in the same distance from the points, so the x-coordinate of R has to be -1. So condition 1 is sufficient,
making the answer A.

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

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

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




_________________

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.

Here in Statement(2),Point R lies on the line y = -3 then the only point on this line y=-3 equidistant from points is (-1,-3),.....so why not D?

From (2) point R can be ANY point on the line y = -3. If it's (-1, -3), then yes, R would be equidistant from points (-3,-3) and (1,-3) but if x-coordinate of point R is anything but -1, then R would NOT be equidistant from points (-3,-3) and (1,-3). For example, (-2, -3), (-1101, -3), ...
_________________

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

Let the co-ordinates of R be (x,y)

(1) The x-coordinate of point R is -1.
x=-1
Distance of R(-1,y) from point (-3,-3) \(= \sqrt{2^2 + (y+3)^2}\)
Distance of R(-1,y) from point (1,-3) \(= \sqrt{2^2 + (y+3)^2}\)
Point R equidistant from points (-3,-3) and (1,-3)
SUFFICIENT

(2) Point R lies on the line y = -3.
y=-3
Distance of R(-1,y) from point (-3,-3) \(= \sqrt{(x+3)^2}=|x+3|\)
Distance of R(-1,y) from point (1,-3) \(= \sqrt{(x-1)^2}=|x-1|\)
Point R is NOT NECESSARILY equidistant from points (-3,-3) and (1,-3)
NOT SUFFICIENT

IMO A

Hello,
In this question I have new dimension to think about

What happens if statement 2 not sufficient but if statement 1 is sufficient and also statement 1 & 2 also sufficient.
In this case, which option to be best possible solution.

Like in this case statement 1 is sufficient
and also statement 1 and 2 is sufficient as statement 1 suggest that x coordinate is -1 and statement 2 suggest that point R lie on the y=-3. The coordinate is (-1, -3) which is also equidistant and hence sufficient.

Hi amiteshiitr,

In DS, the 5 answers choices are always the same 5 options. If you find that either Fact 1 or Fact 2 is Sufficient on its own, then you do NOT need any additional information to answer the question, so Answer C (re: "Together") will not be the correct answer.

Based on your example: Fact 1 is Sufficient; Fact 2 is Insufficient - the final Answer would be Answer A.

GMAT assassins aren't born, they're made,
Rich

Great post. I assume this applies in all cases when you have a perpendicular bisector?
_________________

Yes. Any point on a perpendicular bisector of a segment is equidistant from endpoints of this segment.
_________________

Solution:

Question Stem Analysis:

We need to determine whether point R is equidistant from points (-3,-3) and (1,-3), If it is, then it must be on the perpendicular bisector of the segment connecting (-3,-3) and (1,-3). Since the segment connecting (-3,-3) and (1,-3) is horizontal, the perpendicular bisector of the segment must be vertical, passing through the midpoint of the segment. Since the midpoint of the segment is (-1, -3), the equation of the perpendicular bisector is x = -1. That is, if R is any point on the line x = -1, then it’s equidistant from points (-3,-3) and (1,-3).

Statement One Alone:

Since the x-coordinate of point R is -1, R must be a point on the line x = -1. Therefore, it is equidistant from points (-3,-3) and (1,-3). Statement one alone is sufficient.

Statement Two Alone:

We see that R is a point on the line containing the segment connecting (-3,-3) and (1,-3). However, we can’t determine whether it is equidistant from points (-3,-3) and (1,-3). For example, if R = (-1, -3), i.e., the midpoint of (-3,-3) and (1,-3), then it is equidistant from points (-3,-3) and (1,-3). Otherwise, it is not. Statement two alone is not sufficient.

Answer: A
_________________

Answer: Option A

Video solution by GMATinsight



Get TOPICWISE: Concept Videos | Practice Qns 100+ | Official Qns 50+ | 100% Video solution CLICK.
Two MUST join YouTube channels : GMATinsight (1000+ FREE Videos) and GMATclub

Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1