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

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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, 06:47
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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)
[Reveal] Spoiler: OA

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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, 17:22
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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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

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

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

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

[Reveal] Spoiler:
Attachment:

graph.php.png [ 17.5 KiB | Viewed 10842 times ]

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Re: The area of triangle POQ [#permalink]

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21 Oct 2010, 18:02
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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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Re: Good Question ...Number system [#permalink]

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22 Oct 2010, 22:08
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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

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

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

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

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

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

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23 Oct 2010, 05:35
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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, 21:02
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, 04:30
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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

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

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

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

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, 04:34
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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

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

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

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

But how can we assume that 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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01 Mar 2014, 12:16
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Re: Good Question ...Number system [#permalink]

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05 Apr 2014, 16:28
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
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Re: The area of triangle POQ [#permalink]

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22 Apr 2014, 21:16
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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Re: Good Question ...Number system [#permalink]

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23 Apr 2014, 07:54
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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, 23:59
Hi Bunnel,

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

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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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24 Apr 2014, 02:04
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, 02:26
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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

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

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

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

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

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

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07 May 2014, 02:34
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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

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

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

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

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, 08:11
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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, 09:04
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 $$\frac{OC}{OB}=\frac{OA}{AB}$$ --> $$OC=3$$;

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

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

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

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, 11:27
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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21 Mar 2016, 11:01
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Re: If P and Q are two points on the line 3x + 4y =-15 such that   [#permalink] 21 Mar 2016, 11:01

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