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IMO E

We need to prove \(x^2+y^2 = w^2+z^2\)

Let us consider statement 1 :

|x|+|y|=|w|+|z| => |x|-|z| = |w|-|y|
=> \((|x|-|z|)^2 = (|w|-|y|)^2\)
=> \(|x|^2 + |z|^2 - 2|x||z| = |w|^2 + |y|^2 -2|w||y|\)

So we cannot derive anything from this statement alone.

Now let's use statement 2:
\(x/y = w/z\) => xz = wy

Using both these statements we can say \(|x|^2 + |z|^2=|w|^2 + |y|^2\)

\(|x|^2 - |y|^2 =|w|^2 - |z|^2\), Hence we cannot prove anything even by using both the statements. The only case possible to do so is |x| = |z| & |y| = |w|.
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In a rectangular coordinate system, are the points (x, y) and (w, z) equidistant from the origin?
So, question is asking, is \(x^2 +y^2 = w^2 + z^2 \)?

Stat1: |x|+|y|=|w|+|z|
or, |x|-|z|=|w|-|y| or, \(x^2 +z^2\) -2zx= \(w^2 +y^2\) -2yw. Not sufficient

Stat2: x/y=w/z, so, slope of (x, y) and (0,0) equals to slope of (w, z) and (0,0). But, we don't have an idea for distance. Not sufficient

Combining both, both points (x, y) and (w, z) have same slope and sum of absolute value of co-ordinates are equal too. Which makes points same or mirror image. So, distance from origin will be equal.

So, I think C. :)
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Hello from the GMAT Club BumpBot!

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