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# If P and Q are two points on the line 3x + 4y =-15 such that

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Intern
Joined: 16 Jul 2010
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If P and Q are two points on the line 3x + 4y =-15 such that  [#permalink]

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21 Oct 2010, 05:47
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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)
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Joined: 02 Sep 2009
Posts: 51223
If P and Q are two points on the line 3x + 4y =-15 such that  [#permalink]

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21 Oct 2010, 16:22
8
12
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:

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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

$$CP=CQ=\sqrt{OP^2-OC^2}=\sqrt{9^2-3^2}=6\sqrt{2}$$;

$$PQ=CP+CQ=12\sqrt{2}$$;

$$Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}$$.

P.S. Please post the new questions in separate threads.
Attachment:

graph.php.png [ 17.5 KiB | Viewed 17431 times ]

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Manager
Joined: 30 Sep 2010
Posts: 55
Re: The area of triangle POQ  [#permalink]

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21 Oct 2010, 17:02
4
another way to calculate OC is:
(distance of (a,b) from mx+ny+z=0 is: (ma+nb+z)/sqrt(m^2 + n^2)
SO, OC = distance of O from the line 3x+4y+15= 0 is:

(3 * 0 + 4*0 +15)/sqrt(3^2 + 4^2) = 15/5 = 3
Then we calculate CP=CQ as per Bunel.. and the answer comes out as a) 18* sqrt(2)
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Joined: 26 Mar 2010
Posts: 94
Re: Good Question ...Number system  [#permalink]

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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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

$$CP=CQ=\sqrt{OP^2-OC^2}=\sqrt{9^2-3^2}=6\sqrt{2}$$;

$$PQ=CP+CQ=12\sqrt{2}$$;

$$Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}$$.

P.S. Please post the new questions in separate threads.

Bunuel, what makes us think that point o is Origin in this case???
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Joined: 02 Sep 2009
Posts: 51223
Re: Good Question ...Number system  [#permalink]

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23 Oct 2010, 04:35
1
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.
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Re: Good Question ...Number system  [#permalink]

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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.

Thanks for the info...Bunuel
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Re: Good Question ...Number system  [#permalink]

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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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

$$CP=CQ=\sqrt{OP^2-OC^2}=\sqrt{9^2-3^2}=6\sqrt{2}$$;

$$PQ=CP+CQ=12\sqrt{2}$$;

$$Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}$$.

P.S. Please post the new questions in separate threads.

But how can we assume that O is the origin??
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Re: Good Question ...Number system  [#permalink]

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13 Jul 2012, 03:34
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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

$$CP=CQ=\sqrt{OP^2-OC^2}=\sqrt{9^2-3^2}=6\sqrt{2}$$;

$$PQ=CP+CQ=12\sqrt{2}$$;

$$Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}$$.

P.S. Please post the new questions in separate threads.

But how can we assume that O is the origin??

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Re: Good Question ...Number system  [#permalink]

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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.

Thanks!
Cheers
J
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Joined: 04 Jan 2014
Posts: 104
Re: The area of triangle POQ  [#permalink]

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22 Apr 2014, 20:16
krushna wrote:
another way to calculate OC is:
(distance of (a,b) from mx+ny+z=0 is: (ma+nb+z)/sqrt(m^2 + n^2)
SO, OC = distance of O from the line 3x+4y+15= 0 is:

(3 * 0 + 4*0 +15)/sqrt(3^2 + 4^2) = 15/5 = 3

Hi there, how do we look at m, n and z from the chart?
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Joined: 02 Sep 2009
Posts: 51223
Re: Good Question ...Number system  [#permalink]

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23 Apr 2014, 06:54
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.

Thanks!
Cheers
J

Important property: the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. Check here: if-arc-pqr-above-is-a-semicircle-what-is-the-length-of-144057.html#p1154669

OC is a perpendicular to hypotenuse AB, thus triangles AOB and OCB are similar.

Does this make sense?
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Re: If P and Q are two points on the line 3x + 4y =-15 such that  [#permalink]

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23 Apr 2014, 22:59
Hi Bunnel,

Could you please explain how we can derive SQRT (OA^2 + OB^2) = 25/4?
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Posts: 51223
Re: If P and Q are two points on the line 3x + 4y =-15 such that  [#permalink]

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24 Apr 2014, 01:04
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).

Does this make sense?
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Re: Good Question ...Number system  [#permalink]

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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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

$$CP=CQ=\sqrt{OP^2-OC^2}=\sqrt{9^2-3^2}=6\sqrt{2}$$;

$$PQ=CP+CQ=12\sqrt{2}$$;

$$Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}$$.

P.S. Please post the new questions in separate threads.

Bunuel under which property of similarity would AOB and OCB be similar ? SAS , SSS , ASA ??
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Joined: 02 Sep 2009
Posts: 51223
Re: Good Question ...Number system  [#permalink]

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07 May 2014, 01:34
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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

$$CP=CQ=\sqrt{OP^2-OC^2}=\sqrt{9^2-3^2}=6\sqrt{2}$$;

$$PQ=CP+CQ=12\sqrt{2}$$;

$$Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}$$.

P.S. Please post the new questions in separate threads.

Bunuel under which property of similarity would AOB and OCB be similar ? SAS , SSS , ASA ??

SAS, SSS and ASA conditions are to determine whether the triangles are congruent (same).

The triangles AOB and OCB are similar because they have equal angles. Check here: if-arc-pqr-above-is-a-semicircle-what-is-the-length-of-144057.html#p1154669

Hope it helps.
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Re: If P and Q are two points on the line 3x + 4y =-15 such that  [#permalink]

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27 Aug 2014, 07:11
1
Please post complete question.

O is the origin...
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Re: If P and Q are two points on the line 3x + 4y =-15 such that  [#permalink]

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07 Mar 2015, 08:04
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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

$$CP=CQ=\sqrt{OP^2-OC^2}=\sqrt{9^2-3^2}=6\sqrt{2}$$;

$$PQ=CP+CQ=12\sqrt{2}$$;

$$Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}$$.

P.S. Please post the new questions in separate threads.

Hi,

Can we say that OCA is a primitive pythagorean triple 3:4:5 and deduce that OC =3?

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If P and Q are two points on the line 3x + 4y =-15 such that  [#permalink]

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08 Mar 2015, 10:27
Cannot approach unless it is stated that O is the origin.

If O is not stated as the origin, then we can even take A and B as points and draw an isoscless triangle POQ .

In that case the area will be 25/8 * ( sq.rt 47 * sq.rt 97)
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Re: If P and Q are two points on the line 3x + 4y =-15 such that  [#permalink]

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25 May 2017, 12:52
kobinaot wrote:
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)

Question is good, but nowhere it is mentioned that O is a center i.e. (0,0) co-ordinates.
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Re: If P and Q are two points on the line 3x + 4y =-15 such that  [#permalink]

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13 Aug 2018, 02:13
kobinaot wrote:
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)

Please find the solution as attached.

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Re: If P and Q are two points on the line 3x + 4y =-15 such that &nbs [#permalink] 13 Aug 2018, 02:13
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# If P and Q are two points on the line 3x + 4y =-15 such that

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