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# In the coordinate plane, points (a,b) and (c,d) are

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Intern
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In the coordinate plane, points (a,b) and (c,d) are [#permalink]  09 Aug 2016, 16:11
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|a| > |c|
In the coordinate plane, points (a,b) and (c,d) are equidistant from the origin.

 Quantity A Quantity B |B| |D|

A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

Last edited by GreenlightTestPrep on 02 Feb 2019, 12:54, edited 3 times in total.
Edited the OA
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Re: In the coordinate plane, points (a,b) and (c,d) are [#permalink]  09 Aug 2016, 17:12
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Expert's post
GREhelp wrote:
Manhattan Prep: 5LB Book

In the coordinate plane, points (a,b) and (c,d) are equidistant from the origin. |a| > |c|

Quantity A
|B|

Quantity B
|D|

Here we have 2 points namely X (a,b) and Y (c,d). Now the points are equidistant from origin.

$$a^2 + b^2 = c^2 + d^2$$ ........ 1

Given that

$$|a| > |c|$$ or

$$a^2 > c^2$$

adding$$b^2$$ on both sides

$$a^2 + b^2 > c^2 + b^2$$

from eqn 1

$$c^2 + d^2 > c^2 + b^2$$

or $$d^2 > b^2$$ or $$|d| > |b|$$

Hence option B is correct.
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Re: In the coordinate plane, points (a,b) and (c,d) are [#permalink]  10 Aug 2016, 00:37
Expert's post
Please follow the rules for posting in verbal section. Use the tags, in particular.

Regards
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Re: In the coordinate plane, points (a,b) and (c,d) are [#permalink]  10 Aug 2016, 11:08
Hi Sandy,
Thanks for your response, That was a huge help. I understand how you solved the question and steps taken. However, how did you know when you first looked at the question to do a^2 + b^2 = c^2 + d^2 ????

I understand how you got a^2 > c^2 its because the |a| > |c| which means that A and C can not be negative as a result you did a^2 >c^2. I'm just trying to make sure I understand the underlying concept so I don't make the same mistake again.
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Re: In the coordinate plane, points (a,b) and (c,d) are [#permalink]  10 Aug 2016, 13:02
Expert's post
GREhelp wrote:
Hi Sandy,
Thanks for your response, That was a huge help. I understand how you solved the question and steps taken. However, how did you know when you first looked at the question to do a^2 + b^2 = c^2 + d^2 ????

I understand how you got a^2 > c^2 its because the |a| > |c| which means that A and C can not be negative as a result you did a^2 >c^2. I'm just trying to make sure I understand the underlying concept so I don't make the same mistake again.

Hi Grehelp,

Well actually I solved the problem mentally. I knew that |a| > |c| so |b| < |d| for the two points to be equidistant from origin. This usually comes with a bit of experience and understanding of math over time.

However there is a Hack. Whenever faced with a problem like this one, try and write down all the equations from a problem statement and try to put one equation into another and check if you find some new info. This usually works.

In this case we had only 2 equations. I just tried to combine them to form a new one.

With some practice you can solve these problems easily as well.

Regards,
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Re: In the coordinate plane, points (a,b) and (c,d) are [#permalink]  28 Feb 2018, 14:34
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It's answer should be choice B.
Kindly correct the OA.

Here's how:

Using distance equation:

Distance of point (a,b) from origin(0,0) = (a - 0)^2 + (b - 0)^2
= a^2 + b^2

Similarly,
Distance of point (c,d) from origin(0,0) = (c - 0)^2 + (d - 0)^2
= c^2 + d^2

As, both distances are equal, so

a^2 + b^2 = c^2 + d^2

Now, according to the given condition, absolute value of a is greater than that of d. Thus, in order for making the Left Hand Side (L.H.S) equals to Right Hand Side (R.H.S) of the equation: a^2 + b^2 = c^2 + d^2, we must come to the point that the absolute value of b must always be less that that of d.

So, Quantity b is greater always.

Thus choice B is correct.
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Re: In the coordinate plane, points (a,b) and (c,d) are [#permalink]  09 Mar 2018, 02:39
The answer given to this question is definitely wrong, it should be B, as explained by sandy earlier.
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Re: In the coordinate plane, points (a,b) and (c,d) are [#permalink]  09 Mar 2018, 04:04
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Re: In the coordinate plane, points (a,b) and (c,d) are [#permalink]  09 Mar 2018, 15:38
Expert's post
Done.

Thank you.
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Re: In the coordinate plane, points (a,b) and (c,d) are [#permalink]  02 Feb 2019, 13:13
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GREhelp wrote:
|a| > |c|
In the coordinate plane, points (a,b) and (c,d) are equidistant from the origin.

 Quantity A Quantity B |B| |D|

Here's another approach:

KEY PROPERTIES:
If x² > y² then |x| > |y|
If |x| > |y| then x² > y²

GIVEN: (a,b) and (c,d) are equidistant from the origin.
The origin has coordinates (0, 0)

When we apply the formula for finding the distance between two points, we get:

The distance between (a, b) and (0, 0) = √[(a - 0)² + (b - 0)²]
= √(a² + b²)

The distance between (c, d) and (0, 0) = √[(c - 0)² + (d - 0)²]
= √(c² + d²)

So, we can write: √(a² + b²) = √(c² + d²)
Square both sides to get: a² + b² = c² + d²
Subtract c² from both sides to get: a² + b² - c² = d²
Subtract b² from both sides to get: a² - c² = d² - b²

GIVEN: |a| > |c|
This also tells us that a² > c²
If we subtract c² from both sides we get: a² - c² > 0

Since, we know that a² - c² = d² - b², we can also conclude that d² - b² > 0

Now take d² - b² > 0 and add b² to both sides to get: d² > b²
Finally, if d² > b², then we know that |d| > |b|

Cheers,
Brent
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Re: In the coordinate plane, points (a,b) and (c,d) are   [#permalink] 02 Feb 2019, 13:13
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