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

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

Consider two possible cases below:In both cases TX = 2*PX.

Hope it's clear.

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

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.

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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