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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 ?

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. Thus, C is the answer.

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.

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.

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

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!

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.
_________________

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.

Sent from my SM-N910H using Tapatalk

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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