Tough DS

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Tough DS

by rainmaker » Sat Aug 22, 2009 6:51 pm
Please explain your answer:

In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

(2) sqrt(a^2) + sqrt(b^2) = sqrt(c^2) + sqrt(d^2)

where ^ means "raised to the power of"
Source: — Data Sufficiency |

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by iamjakekim » Sat Aug 22, 2009 10:08 pm
I think it is D

STM1.

if you cross multiply, you get b/a = d/c

STM2.

We need to know whether distance is equal.

For example, (1,4) and (4,1) are equal distance.
(-1,5) and (1,-5) are equal distance.!!

DDDDDD

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by shanrizvi » Sun Aug 23, 2009 1:48 pm
In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

Statement 1: a/b = c/d

If (a,b) is (2,4) and (c,d) is (-2,-4), a/b = c/d and the points are equidistant. However, if (a,b) is (2,4) and (c,d) is (4,8), a/b = c/d but the points are not equidistant. Hence, the statement is insufficient.

Statement 2: sqrt(a^2) + sqrt(b^2) = sqrt(c^2) + sqrt(d^2)

For any point (x,y), the distant from the origin is sqrt(x^2+y^2). As this is NOT equal to sqrt(a^2) + sqrt(b^2), I say this statement is insufficient.

To prove this, plot any point (x,y) on the graph and consider the triangle bound by lines X=x, Y=y and the line from origin to (x,y). The distance to origin is the hypotenuse of the triangle with the other 2 sides of lengths x and y.

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by pradeepsarathy » Tue Aug 25, 2009 7:07 pm
IMO C

Stmt 1 - a/b = c/d

Case 1 - a = 1, b = 1, c = 2, d = 2
while a/b = c/d, the points are not equidistant from origin

Case 2 - a = -1, b = -1, c = 1, d = 1
a/b = c/d and also the points are equidistant from origin,
Hence stmt 1 alone is insufficient.
Eliminate answer choices A and D

Stmt 2 - sqrt(a^2) + sqrt(b^2) = sqrt(c^2) +sqrt(d^2)

Case 1 - a = 1, b = 2, c = 0, d = 3
while the points satisfy stmt 2, they are not equidistant from origin.

Case 2 - a = 1, b = 2, c = -1, d = -2
the points both satisfy the stmt 2 and also are equidistant from origin.

Hence stmt 2 alone is insufficient

Eliminate choice B.

Combining both stmt 1 and stmt 2 -
we see that the points have to be symmetrically oppisite to each other, and hence they will be equidistant from the origin as well.

Hence both stmts are sufficient.