SajjadAhmad wrote:

In the figure below, which of the following points is three times as far from P as from Q?

(A) (0, 3)

(B) (1, 1)

(C) (4, 5)

(D) (2, 3)

(E) (4, 1)

Nice question. In the case of unit graphs, we can trust our eyes on the really obvious incorrect answers.

1. I started there, and eliminated (A), (C), and (D) in about 10 seconds.

(A) at (0, 3) is closer to P than to Q. Incorrect.

(C) at (4, 5) looks equidistant (and I didn't do the math, just counted: it is up 1 and over 2 from P, over 1 and up 2 for Q) -- at the least, it's certainly not three times as far from P as it is from Q.

(D) at (2, 3) also looks equidistant (it's exactly one unit square diagonal away from both P and Q); again, at the least, it's not three times farther from P than from Q

2. To evaluate (B) and (E), I used the Pythagorean theorem. And I counted unit squares to get lengths.

(B) at (1, 1) is

3 units away from

P.

(B) is 2 horizontal units away from Q (length of one triangle leg), and 1 vertical unit away from Q (length of the other leg). These two legs form a right triangle, so \(a^2 + b^2 = c^2\), where c is the distance from Q.

\(1^2 + 2^2 = distance^2\), or 5 = distance squared, so

\(\sqrt{5}\)=

distance from Q. The latter is about 2.2. Compare Q's distance of 2.2 with P's distance of 3. (B) isn't far enough away from P. Reject. And hope the next one works.

3.

(E) at (4, 1) is 3 horizontal units away from P, and 3 vertical units away from P. Again, Pythagorean theorem for distance from P:

\(3^2 + 3^2 = D^2\), so

\(18 = D^2\), and D =\(\sqrt{18}\), which simplifies to \(3\sqrt{2}\)

(E) is 1 vertical and 1 horizontal unit away from Q.

That's a 45-45-90 right triangle with sides in ratio \(x : x : x\sqrt{2}\).

Legs = 1, so hypotenuse \(= (1)(\sqrt{2}\)). Distance from Q is \(\sqrt{2}\).

P's distance from (E) is \(3\sqrt{2}\), exactly three times that of Q's distance from (E). Correct

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