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25 Dec 2020, 01:22
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E
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Difficulty:
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79% (00:46) correct
21%
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wrong
based on 114
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What are the coordinates of the midpoint of line segment AB in xy-plane?
(1) Line seg AB has length 6
(2) The coordinates of A are (2,4)
(1) Line seg AB has length 6
(2) The coordinates of A are (2,4)
27 Dec 2020, 06:58
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Hoozan
The mid pint of a line segment between 2 points is \(\frac{x_1 + x_2}{2}\) and \(\frac{y_1 + y_2}{2}\)
Statement 1: Line seg AB has length 6
The distance between 2 points X \((x_1,y_1) and B (x_2,y_2)\) is given by
\(XY = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }\)
Since the length is 6, then \(6 = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }\) or \(36 = (x_2 – x_1)^2 + (y_2 – y_1)^2\)
We can substitute values here for \((x_2 - x_1)^2\) and \((y_2 - y_1)^2\)
If \((x_2 - x_1)^2 = 18\) and \((y_2 - y_1)^2\) then \((x_2 - x_1) = (y_2 - y_1) \sqrt{18} = 3\sqrt{2}\).
Then we can have the x1 as \(2\sqrt{2}\) and x2 as \(5\sqrt{2}\) and y1 = \(2\sqrt{2}\) and y2 as \(5\sqrt{2}\)
The mid point would be \(\frac{2\sqrt{2} + 5\sqrt{2}}{2} = 3.5\sqrt{2}\) and \(\frac{2\sqrt{2} + 5\sqrt{2}}{2} = 3.5\sqrt{2}\)
We can also have x1 as \(4\sqrt{2}\) and x2 as \(7\sqrt{2}\) and y1 = \(4\sqrt{2}\) and y2 as \(7\sqrt{2}\)
The mid point would be \(\frac{4\sqrt{2} + 7\sqrt{2}}{2} = 5.5\sqrt{2}\) and \(\frac{4\sqrt{2} + 7\sqrt{2}}{2} = 5.5\sqrt{2}\)
We can get many such values and get that the length of the line segment = 6
Therefore Statement 1 Alone is Insufficient. Answer options could be B, C or E
Statement 2: The co ordinates of A are (2, 4)
Here we do not know the length of the line segment. Then B can take an infinite set of values and we can get an infinite number of values of the mid point.
For eg. If B = (4,2), then the mid point would be x = (2 + 4)/2 = 3 and y = (4 + 2)/2 = 3
If B = (6,4), then the mid point would be x = (2 + 6)/2 = 4 and y = (4 + 4)/2 = 4
Therefore Statement 2 Alone is Insufficient. Answer Options could be C or E
Combining Both Statements:. Length of line segment = 6 and A's co ordinates are (2,4)
Often we might think that this is sufficient.
We know that \((x_2 – x_1)^2 + (y_2 – y_1)^2 = 36\) and putting x1 and y1 as 2 and 4, we get \((x_2 – 2)^2 + (y_2 – 4)^2 = 36\)
Now for different values of \((x_2 – 2)^2\) and \((y_2 – 4)^2\), we will get different values of x1 and x2
Therefore Both Statements together are Insufficient.
Option E
Arun Kumar
10 Oct 2023, 21:14
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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