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Re: Good Question ...Number system [#permalink]
21 Oct 2010, 16:22

3

This post received KUDOS

Expert's post

5

This post was BOOKMARKED

kobinaot wrote:

help pls!

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

Attachment:

graph.php.png [ 17.5 KiB | Viewed 3830 times ]

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

OA=5 and OB=\frac{15}{4} (points A and B are intersection of the line 3x+4y=-15 with the X and Y axis respectively) --> AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4};

As AOB and OCB are similar then \frac{OC}{OB}=\frac{OA}{AB} --> OC=3;

Re: Good Question ...Number system [#permalink]
22 Oct 2010, 21:08

Bunuel wrote:

kobinaot wrote:

help pls!

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

Attachment:

graph.php.png

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

OA=5 and OB=\frac{15}{4} (points A and B are intersection of the line 3x+4y=-15 with the X and Y axis respectively) --> AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4};

As AOB and OCB are similar then \frac{OC}{OB}=\frac{OA}{AB} --> OC=3;

Re: Good Question ...Number system [#permalink]
23 Oct 2010, 04:35

1

This post received KUDOS

Expert's post

utin wrote:

Bunuel, what makes us think that point o is Origin in this case???

Well, we must know the coordinates of the point O to solve the question. As origin in coordinate geometry usually marked with letter O (as the center of a circle) then we can assume that O is origin in this case. Though I think on GMAT it would be explicitly stated. _________________

Re: Good Question ...Number system [#permalink]
23 Oct 2010, 20:02

Bunuel wrote:

utin wrote:

Bunuel, what makes us think that point o is Origin in this case???

Well, we must know the coordinates of the point O to solve the question. As origin in coordinate geometry usually marked with letter O (as the center of a circle) then we can assume that O is origin in this case. Though I think on GMAT it would be explicitly stated.

Re: Good Question ...Number system [#permalink]
13 Jul 2012, 03:30

Bunuel wrote:

kobinaot wrote:

help pls!

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

Attachment:

graph.php.png

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

OA=5 and OB=\frac{15}{4} (points A and B are intersection of the line 3x+4y=-15 with the X and Y axis respectively) --> AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4};

As AOB and OCB are similar then \frac{OC}{OB}=\frac{OA}{AB} --> OC=3;

Re: Good Question ...Number system [#permalink]
13 Jul 2012, 03:34

Expert's post

MacFauz wrote:

Bunuel wrote:

kobinaot wrote:

help pls!

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

Attachment:

graph.php.png

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

OA=5 and OB=\frac{15}{4} (points A and B are intersection of the line 3x+4y=-15 with the X and Y axis respectively) --> AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4};

As AOB and OCB are similar then \frac{OC}{OB}=\frac{OA}{AB} --> OC=3;

Re: If P and Q are two points on the line 3x + 4y =-15 such that [#permalink]
01 Mar 2014, 11:16

Hello from the GMAT Club BumpBot!

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Re: Good Question ...Number system [#permalink]
05 Apr 2014, 15:28

Bunuel could you please elaborate a little bit more on the relation between similar triangles? How could you get OC/OB = OA/AB, I'm not able to see it.

Re: Good Question ...Number system [#permalink]
23 Apr 2014, 06:54

Expert's post

jlgdr wrote:

Bunuel could you please elaborate a little bit more on the relation between similar triangles? How could you get OC/OB = OA/AB, I'm not able to see it.

Re: If P and Q are two points on the line 3x + 4y =-15 such that [#permalink]
24 Apr 2014, 01:04

Expert's post

pretzel wrote:

Hi Bunnel,

Could you please explain how we can derive SQRT (OA^2 + OB^2) = 25/4?

AB is the hypotenuse of right triangle AOB, hence AB=hypotenuse=\sqrt{OA^2+OB^2}, where OA=5 and OB=\frac{15}{4} (points A and B are intersection of the line 3x+4y=-15 with the X and Y axis respectively).

Re: Good Question ...Number system [#permalink]
07 May 2014, 01:26

Bunuel wrote:

kobinaot wrote:

help pls!

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

Attachment:

graph.php.png

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

OA=5 and OB=\frac{15}{4} (points A and B are intersection of the line 3x+4y=-15 with the X and Y axis respectively) --> AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4};

As AOB and OCB are similar then \frac{OC}{OB}=\frac{OA}{AB} --> OC=3;

Re: Good Question ...Number system [#permalink]
07 May 2014, 01:34

Expert's post

himanshujovi wrote:

Bunuel wrote:

kobinaot wrote:

help pls!

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

Attachment:

graph.php.png

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

OA=5 and OB=\frac{15}{4} (points A and B are intersection of the line 3x+4y=-15 with the X and Y axis respectively) --> AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4};

As AOB and OCB are similar then \frac{OC}{OB}=\frac{OA}{AB} --> OC=3;

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