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Math Expert V
Joined: 02 Sep 2009
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On the x-y coordinate grid, are points A(m, n) and B(r, t) equidistant  [#permalink]

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Question Stats: 83% (01:20) correct 17% (00:51) wrong based on 75 sessions

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On the x-y coordinate grid, are points A(m, n) and B(r, t) equidistant from the origin?

(1) |m| = |n| and |r| = |t|
(2) |m| = |r| and |n| = |t|

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On the x-y coordinate grid, are points A and B equidistant from the origin?
(1) |x| = |y| for point A and |x| = |y| for point B
(2) |x|, |y| of point A = |x|, |y| of point B

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Re: On the x-y coordinate grid, are points A(m, n) and B(r, t) equidistant  [#permalink]

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1
Bunuel wrote:
On the x-y coordinate grid, are points A(m, n) and B(r, t) equidistant from the origin?

(1) |a| = |b| and |r| = |t|
(2) |m| = |r| and |n| = |t|

There's a typo in Statement 1; I assume it means to say |m| = |n|, and not "|a| = |b|" (since there is no a or b in the question).

Statement 1 is not sufficient, since it's possible A and B are the same point, and are thus equidistant from the origin, but it's possible A is (1, 1) and B is (1000, 1000) and are at very different distances from the origin.

Statement 2 is sufficient, since if |m| = |r|, then m^2 = r^2, and similarly n^2 = t^2. We use Pythagoras to find the distance from a point to the origin, so the distance from (0, 0) to (m, n) is just √(m^2 + n^2). If m^2 = r^2 and n^2 = t^2, then √(m^2 + n^2) must equal √(r^2 + t^2), the distance from (0, 0) to (r, t).
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Math Expert V
Joined: 02 Sep 2009
Posts: 65012
Re: On the x-y coordinate grid, are points A(m, n) and B(r, t) equidistant  [#permalink]

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IanStewart wrote:
Bunuel wrote:
On the x-y coordinate grid, are points A(m, n) and B(r, t) equidistant from the origin?

(1) |a| = |b| and |r| = |t|
(2) |m| = |r| and |n| = |t|

There's a typo in Statement 1; I assume it means to say |m| = |n|, and not "|a| = |b|" (since there is no a or b in the question).

Statement 1 is not sufficient, since it's possible A and B are the same point, and are thus equidistant from the origin, but it's possible A is (1, 1) and B is (1000, 1000) and are at very different distances from the origin.

Statement 2 is sufficient, since if |m| = |r|, then m^2 = r^2, and similarly n^2 = t^2. We use Pythagoras to find the distance from a point to the origin, so the distance from (0, 0) to (m, n) is just √(m^2 + n^2). If m^2 = r^2 and n^2 = t^2, then √(m^2 + n^2) must equal √(r^2 + t^2), the distance from (0, 0) to (r, t).

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Edited. Thank you.
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On the x-y coordinate grid, are points A(m, n) and B(r, t) equidistant  [#permalink]

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2
Bunuel wrote:
On the x-y coordinate grid, are points A(m, n) and B(r, t) equidistant from the origin?

(1) |m| = |n| and |r| = |t|
(2) |m| = |r| and |n| = |t|

Project PS Butler

On the x-y coordinate grid, are points A and B equidistant from the origin?
(1) |x| = |y| for point A and |x| = |y| for point B
(2) |x|, |y| of point A = |x|, |y| of point B

The distances for A and B to the origin would be $$\sqrt{m^2 + n^2}$$ and $$\sqrt{r^2 + t^2}$$. To have these values equal we must need $$m^2 + n^2 = r^2 + t^2$$. Then we are asking if this equality is true.

Statement 1:
We can reduce the question to "$$2m^2 = 2r^2$$?" Insufficient.

Statement 2:
Plug in to get the left side is $$r^2 + t^2$$, then both sides must be equal. Sufficient.

Ans: B
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Re: On the x-y coordinate grid, are points A(m, n) and B(r, t) equidistant  [#permalink]

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Attachments 1.PNG [ 42.21 KiB | Viewed 462 times ] Re: On the x-y coordinate grid, are points A(m, n) and B(r, t) equidistant   [#permalink] 25 May 2020, 19:33

# On the x-y coordinate grid, are points A(m, n) and B(r, t) equidistant  