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geometry-3 slight tricky question

Expert replies
Source: — Problem Solving |

by anuprajan5 » Tue Oct 23, 2012 10:37 pm
The question rephrased says that if point A is present, which is 2/3 the distance from Q, what is the co-ordinates of A.

Just from a visual, I would skip C and D.

Distance between P and Q is root 18 ( root((3-0)^2+(2-(-1))^2)) which equals 3 root 2.

2/3 distance of that is 2 root 2.

[spoiler]Taking choice B distance between P and A is root(4+4) = 2 root 2[/spoiler]
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by GMATGuruNY » Wed Oct 24, 2012 2:07 am
Image

In the figure above, the point on segment PQ that is twice as far from P as from Q is

A) (3,1)

B) (2,1)

C) (2,-1)

D) (1.5,0.5)

E) (1,0)

Can anyone solve this in detail?
Don't calculate; use the answer choices.

The x coordinate of P is 0.
The x coordinate of Q is 3.
The midpoint between P and Q has an x coordinate of (0+3)/2 = 1.5
Thus, the x coordinate of a point closer to Q must be between 1.5 and 3.
Eliminate A, D, and E.
In answer choices B and C, x=2.
Since at x=2 line segment PQ is above the x axis, the y coordinate must be positive.
Eliminate C.

The correct answer is B.
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by gmat6087 » Wed Oct 24, 2012 2:09 am
'manpreet singh wrote:Image

kindly explain the answer?
Another approach.

total X distance 3-0=3
Total Y distance 2-(-1)=3

point lies on line such that it is twice the distance from p than from Q

ratio=2:1
divide the line into 3 equal parts:
x: division: 3/3=1 (each x segment)---eq1
Y: division: 3/3=1 (each Y segment)---eq2

2 parts from x segment= 1*2=>0+2=2 from eq1
2 parts from y segment= 1*2=>-1+1+1=1 from eq2(since y starts from -1)
so point is (2,1)
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by \'manpreet singh » Wed Oct 24, 2012 9:11 pm
Thanks guys,I guess i got puzzled with statement and yes answer is indeed B

Singh
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