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Re: The coordinates of points A and B are p, q and (r, s). Is |q| > |s|? [#permalink]
Bunuel wrote:
The coordinates of points A and B are (p, q) and (r, s). Is |q| > |s|?

(1) The points A and B are equidistant from the origin.
(2) |p| > |r|.


if the two points A and B are equidistant then lets say point A (0,4) and point B (4,0) then |q| >|s|?

Pls explain thanks
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Re: The coordinates of points A and B are p, q and (r, s). Is |q| > |s|? [#permalink]
We have two points A(p,q) and B(r,s) on the co-ordinate plane. Note that these points can be in any of the four quadrants and hence pay due attention to the signs of the co-ordinates as well.

In the question, we are trying to find if the absolute value of q is more than the absolute value of s.

Statement II alone does not give us any information about the question we are trying to answer, since it only talks about p and r. Therefore, let us eliminate options B and D since statement II alone is insufficient.
The possible answer options at this stage are A, C or E.

From statement I alone, we only know that the two points are equidistant from the origin.

If we take A(3,4) and B(4,3), both points are equidistant from the origin. In this case q = 4 and s = 3 and hence |q| > |s|.

If we take A(1,1) and B(-1,-1), both points are equidistant from the origin, but |q| = |s|.

Statement I alone is insufficient. Answer option A can be eliminated, possible answer options are C or E.

Combining statements I and II, we know that A and B are equidistant from the origin, and |p| > |r|.

Since A and B are equidistant from the origin, \(p^2 + q^2 = r^2 + s^2\). Also, distance of p from ZERO on x-axis is higher than the distance of r from ZERO on the x-axis (remember that p and r are the x co-ordinates).
If the equation above has to be satisfied, then distance of q from ZERO on y-axis should be lower than the distance of s from ZERO on y-axis. This is to balance out the equation. Otherwise, the LHS would become greater than the RHS.

Another easy way of understanding this concept is by looking at co-ordinates like (4,3) and (3,4). Here since |p|>|r|, you can observe that |q|<|s|. If you take the case of (1,1) and (-1,-1), you see that |p| = |r| and hence it turns out that |q| = |s| so that the equation holds.

The combination of statements is sufficient to say |q| < |s| and answer the question with a definite NO. Answer option E can be eliminated.
The correct answer option is C.

Hope that helps!
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Re: The coordinates of points A and B are p, q and (r, s). Is |q| > |s|? [#permalink]
zanaik89 wrote:
Bunuel wrote:
The coordinates of points A and B are (p, q) and (r, s). Is |q| > |s|?

(1) The points A and B are equidistant from the origin.
(2) |p| > |r|.


if the two points A and B are equidistant then lets say point A (0,4) and point B (4,0) then |q| >|s|?

Pls explain thanks


Hello Zanaik,

You cannot prove or disprove a statement based on one set of values alone. What if we take points A(4,3) and B(3,4)? These are equidistant from the origin as well. Here |q| < |s|. If you swap the co-ordinates, you will also be able to obtain |q| > |s|.

If you take (1,1) and (-1,-1), you will be able to obtain |q| = |s| also.

And that is WHY, statement I alone is insufficient. So, it's not about proving that |q| > |s|, you are better off disproving it to prove that the statement is insufficient.

Hope that helps!
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Re: The coordinates of points A and B are p, q and (r, s). Is |q| > |s|? [#permalink]
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Re: The coordinates of points A and B are p, q and (r, s). Is |q| > |s|? [#permalink]
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