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If P and Q are two points on the line 3x + 4y =-15 such that 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)
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A
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If P and Q are two points on the line 3x + 4y =-15 such that 21 Oct 2010, 17:22
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kobinaot
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)
Show more

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

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

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

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

Answer: A.

P.S. Please post the new questions in separate threads.
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If P and Q are two points on the line 3x + 4y =-15 such that 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 22 Oct 2010, 22:08
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kobinaot
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.
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Bunuel, what makes us think that point o is Origin in this case???
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If P and Q are two points on the line 3x + 4y =-15 such that 23 Oct 2010, 05:35
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utin
Bunuel, what makes us think that point o is Origin in this case???
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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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If P and Q are two points on the line 3x + 4y =-15 such that 23 Oct 2010, 21:02
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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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Thanks for the info...Bunuel :)
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If P and Q are two points on the line 3x + 4y =-15 such that 13 Jul 2012, 04:30
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Bunuel
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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.
Show more

But how can we assume that O is the origin??
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Bunuel
kobinaot
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??
Show more

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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If P and Q are two points on the line 3x + 4y =-15 such that 05 Apr 2014, 16:28
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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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If P and Q are two points on the line 3x + 4y =-15 such that 22 Apr 2014, 21:16
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krushna
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
Show more

Hi there, how do we look at m, n and z from the chart?
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If P and Q are two points on the line 3x + 4y =-15 such that 23 Apr 2014, 07:54
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jlgdr
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
Show more

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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If P and Q are two points on the line 3x + 4y =-15 such that 23 Apr 2014, 23:59
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Hi Bunnel,

Could you please explain how we can derive SQRT (OA^2 + OB^2) = 25/4?
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If P and Q are two points on the line 3x + 4y =-15 such that 24 Apr 2014, 02:04
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Hi Bunnel,

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

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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If P and Q are two points on the line 3x + 4y =-15 such that 07 May 2014, 02:26
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Bunuel
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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.
Show more

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

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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If P and Q are two points on the line 3x + 4y =-15 such that 27 Aug 2014, 08:11
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Please post complete question.

O is the origin...
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Bunuel
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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.
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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 08 Mar 2015, 11:27
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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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If P and Q are two points on the line 3x + 4y =-15 such that 25 May 2017, 13:52
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kobinaot
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)
Show more


Question is good, but nowhere it is mentioned that O is a center i.e. (0,0) co-ordinates.
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If P and Q are two points on the line 3x + 4y =-15 such that 13 Aug 2018, 03:13
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kobinaot
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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Please find the solution as attached.

Answer: Option A
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