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by leumas » Fri Sep 09, 2011 10:15 pm
Line segment PQ in a graph. P (0,-1) ends at Q (3,2). In between there are two points (1,0) and (2,1).



Question:
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)
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by cans » Fri Sep 09, 2011 10:28 pm
P(0,-1) Q(3,2)
twice far from P as from Q is: (2:1)
x: (2*3 + 1*0)/3 = 2
y: (2*2 + 1*(-1))/3 = 1
IMO B
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by leumas » Fri Sep 09, 2011 10:33 pm
cans wrote:P(0,-1) Q(3,2)
twice far from P as from Q is: (2:1)
x: (2*3 + 1*0)/3 = 2
y: (2*2 + 1*(-1))/3 = 1
IMO B
Can you please explain the logic! I couldn't get it.
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by sl750 » Fri Sep 09, 2011 10:40 pm
Or you can say, rise 2 units up the y-axis and run 1 unit along the x-axis to the right, so it is (2,1)
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by GMATGuruNY » Sat Sep 10, 2011 2:30 am
I posted a solution here:

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