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# In the xy-plane, point O is located at the origin, point A has coordin

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In the xy-plane, point O is located at the origin, point A has coordin [#permalink]

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02 Jul 2016, 09:38
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In the xy-plane, point O is located at the origin, point A has coordinates (p,q), and point B has coordinates (r,0). If p, q, and r are all positive values and AO > AB, is the area of triangular region ABO less than 12 ?

(1) r = 7

(2) p = 4 and q = 3
[Reveal] Spoiler: OA

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Re: In the xy-plane, point O is located at the origin, point A has coordin [#permalink]

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02 Jul 2016, 09:51
I am getting D.

St 2 is correct for obvious reasons.
St 1 Given r=7
Also, AO>AB

[(P-0)^2+(Q-0)^2]^1/2>[(P-R)^2+Q^2]^1/2
P^2+Q^2>(P-R)^2+Q^2
P^2>(P-R)^2
P^2>P^2+R^2-2PR
2PR>R^2
2P>R
2P>7
P>3.5
Area(AOB)= 1/2b.h
=1/2X7XP
since P>3.5
Area(AOB)>12.25
Thus A is sufficient.

Let me know what mistake I made..

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Re: In the xy-plane, point O is located at the origin, point A has coordin [#permalink]

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02 Jul 2016, 10:16
Saurabh I think you considered the height of the triangle in incorrectly

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Re: In the xy-plane, point O is located at the origin, point A has coordin [#permalink]

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02 Jul 2016, 10:20
By using Heron's Formula
Area of triangle is 1/2(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))
Consider statement 1 which is not enough
Now consider statement 2 which still does not gives us all three coordinates of vertices.

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Re: In the xy-plane, point O is located at the origin, point A has coordin [#permalink]

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02 Jul 2016, 11:56
comprehensive soln anyone folks?

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Re: In the xy-plane, point O is located at the origin, point A has coordin [#permalink]

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07 Jul 2016, 23:31
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Ans should be B.
The most important clue that we have is AO>AB
This restricts how far r can go.
AO from coordinates will be 5 units in length. So maximum value of AB < 5.
And all points are in first quadrant only so maximum value of r can be 8 (excluding 8).
So maximum area of tringle could be
1/2 * 8 * 3 = 12 (excluding 12)
Let me know if its not clear or if I am wrong somewhere.

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Re: In the xy-plane, point O is located at the origin, point A has coordin [#permalink]

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07 Jul 2016, 23:56
To add to my solution above,
Algebraically we can show how r should be less than 8.
We know that length of a line between two coordinates is equal to sqrt of ( (x1-x2)^2 + (y1-y2)^2 )
So from AO > AB we get:
AO^2 > AB^2 (since both sides are positive )
So (4-0)^2 + (3-0)^2 > (r-4)^2 + (3-0)^2
Simplifying :
(r-4)^2 < 16
This gives us 0 < r < 8

Let me know if its not clear or if I am wrong somewhere.

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Re: In the xy-plane, point O is located at the origin, point A has coordin [#permalink]

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25 Oct 2016, 07:20
saurabh87 wrote:
In the xy-plane, point O is located at the origin, point A has coordinates (p,q), and point B has coordinates (r,0). If p, q, and r are all positive values and AO > AB, is the area of triangular region ABO less than 12 ?

(1) r = 7

(2) p = 4 and q = 3

oh wow...took 3 minutes just to make sure i don't need 1...
a truly 700 level question

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Re: In the xy-plane, point O is located at the origin, point A has coordin [#permalink]

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15 Jul 2017, 07:28
shashanksagar wrote:
Ans should be B.
The most important clue that we have is AO>AB
This restricts how far r can go.
AO from coordinates will be 5 units in length. So maximum value of AB < 5.
And all points are in first quadrant only so maximum value of r can be 8 (excluding 8).
So maximum area of tringle could be
1/2 * 8 * 3 = 12 (excluding 12)
Let me know if its not clear or if I am wrong somewhere.

Sent from my SM-N910H using Tapatalk

Hey! Why is it excluding 8 here? maximum value of r can be 8 rite? area can also be 12, which is the maximum. Pls throw some light here!

Thanks,
Uma

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In the xy-plane, point O is located at the origin, point A has coordin [#permalink]

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15 Jul 2017, 09:01
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Expert's post
saurabh87 wrote:
In the xy-plane, point O is located at the origin, point A has coordinates (p,q), and point B has coordinates (r,0). If p, q, and r are all positive values and AO > AB, is the area of triangular region ABO less than 12 ?

(1) r = 7

(2) p = 4 and q = 3

Detailed solution.

We can clearly see the following things from the question statement:

1. All p,q,r >0
2. As AO > AB

$$p^2 + q^2 > (p-r)^2 + q^2$$

Giving you, $$2*p*r > r^2$$

2 cases possible -->

Case 1: $$2pr-r^2 > 0$$ ---> r>0 and 2p>r or

Case 2:$$2pr-r^2 >0$$ ---> r<0 and 2p<r . BUT this goes against #1 above as r MUST BE > 0. Hence ignore this case.

Thus the only possible set of values are r>0 and 2p>r

3. Area of Triangle ABO = 0.5*r*q

The question asks, is 0.5*r*q<12 or is r*q<24

Per Statement 1: No information on q. NOT SUFFICIENT.

Per Statement 2: q=3 and p=2 --> 2p>r --> r<8

Thus, the value of r*q MUST BE <24 as q=3 and r<8. By any combination of values of r and q, you will NEVER get a value >= 24.

Hence, this statement is sufficient.

Hope this helps. The question is very straightforward if you break it down into recognizable chunks.
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Re: In the xy-plane, point O is located at the origin, point A has coordin [#permalink]

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20 Aug 2017, 22:20
shashanksagar wrote:
Ans should be B.
The most important clue that we have is AO>AB
This restricts how far r can go.
AO from coordinates will be 5 units in length. So maximum value of AB < 5.
And all points are in first quadrant only so maximum value of r can be 8 (excluding 8).
So maximum area of tringle could be
1/2 * 8 * 3 = 12 (excluding 12)
Let me know if its not clear or if I am wrong somewhere.

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While I understood the algebraic approach to deduce that r should be <8. The length of the 3rd side should be: greater than 1 and less than 9. Is there any other way to deduce that r<8?

Bunuel: Please throw some light here.
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Re: In the xy-plane, point O is located at the origin, point A has coordin   [#permalink] 20 Aug 2017, 22:20
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