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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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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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Re: The area of triangle POQ [#permalink]

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

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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:
Image
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 \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

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

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

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

Answer: A.

P.S. Please post the new questions in separate threads.
[Reveal] Spoiler:
Attachment:
graph.php.png
graph.php.png [ 17.5 KiB | Viewed 11348 times ]

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

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

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Please post complete question.

O is the origin...
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Re: Good Question ...Number system [#permalink]

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New post 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 \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

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

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

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

Answer: A.

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

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New post 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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New post 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 \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

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

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

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

Answer: A.

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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New post 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 \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

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

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

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

Answer: A.

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


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


Please read the thread completely: if-p-and-q-are-two-points-on-the-line-3x-4y-15-such-that-103360.html#p805599
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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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Re: Good Question ...Number system [#permalink]

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New post 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!
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Re: The area of triangle POQ [#permalink]

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New post 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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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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New post 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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New post 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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New post 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 \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

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

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

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

Answer: A.

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

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New post 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 \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

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

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

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

Answer: A.

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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New post 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 \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

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

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

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

Answer: A.

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?

Thanks in advance :-)

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

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New post 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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