In figure below P and T lie on line L. How many different

Question

Geometry Line problem.JPG In figure below P and T lie on line L. How many different points on L are twice as for from point T as from point P? (A) Infinite number (B) 4 (C) 2 (D) 1 (E) None

Bunuel · Accepted Answer

https://gmatclub.com/forum/download/file.php?id=36235 In the figure above, points P and T lie on line L. How many different points on L are twice as far from point T as from point P? (A) 6 (B) 4 (C) 3 (D) 2 (E) 1 2017-07-13_1028.png

srini123 · Answer

2

1/3 of PT
and PT left to P

Magdak · Answer

Banuel, I didn't understand the question. Could you pls. visualize the solution on the line? Thanks.

Bunuel · Answer

Consider two possible cases below: Untitled.png In both cases TX = 2*PX. Hope it's clear.

shashankism · Answer

We are searching for points on L which is twice as far from P than from point T 
means, P is the midpoint of T and the points which are being searched.

There is only one such possibility on line L.

Answer E

generis · Answer

I think the answer is D.

I assigned numbers.

After fooling around with the numbers, it was easier for me to use literal thirds, as in,  \frac{1}{3}

FIRST SCENARIO: Let P = 0 and T = 1

________P___________________T
________0___________________1

Imagine A is the point twice as far from T as from P. First place A can be is BETWEEN P and T. A =  \frac{1}{3}

____P____A______________T
____0____ \frac{1}{3} _____________1

Distance from P to A:  \frac{1}{3}

Distance from T to A:  \frac{2}{3}

SCENARIO 2: 
The second possible point for A is to the left of P.

Let T =  \frac{1}{3} , P = 0, and A = - \frac{1}{3} .

__A________P________T
_- \frac{1}{3}  _______0________ \frac{1}{3}

Distance from P to A:  \frac{1}{3}

Distance from T to A:  \frac{2}{3}

That's all I can come up with. If point A lies to the right of T, I cannot construct any numbers that will work.

My intuition tells me that there is a problem with absolute value. If T is origin, and distance point is to the right of T, no point to the right of T will yield a distance twice as far from T as P.

Answer D

longhaul123 , this should help with your question.

(  Bharath99 , and  shashankism  -- did you deploy some concept to which I do not refer here? I don't see any actual numbers, so I'm curious about your reasoning.)

Whew. I hope that's right, and I hope it helps.

JeffTargetTestPrep · Answer

Let’s place a point Q on line L, such that point Q is twice as far from point T as it is from point P. We see that we can place Q somewhere i) to the left of P, ii) in between P and T, or iii) to the right of T. We need to see which of these 3 ways will make Q twice as far from point T as from point P.

To make things easier, let’s assume line L is a number line, and point P is at the number 3 and point T is at the number 6.

If we place point Q to the left of P, we can place it on the number 0 so that QP = |3 - 0| = 3 and QT = |6 - 0| = 6. We see that Q is twice as far from point T as from point P.

If we place point Q in between P and T, we can place it on the number 4 so that QP = |3 - 4| = 1 and QT = |6 - 4| = 2. We see that Q is also twice as far from point T as from point P.

If we place point Q to the right of T, there is no way we can make Q twice as far from point T as from point P, since Q is closer to T than to P (i.e., the distance between Q and T is always less than the distance between Q and P).

Thus, we have two locations to place Q such that Q is twice as far from point T as from point P.

Answer: D

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