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If P and Q are two points on the line 3x + 4y =15 such that
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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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If P and Q are two points on the line 3x + 4y =15 such that
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21 Oct 2010, 16:22
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 \(\frac{OC}{OB}=\frac{OA}{AB}\) > \(OC=3\); \(CP=CQ=\sqrt{OP^2OC^2}=\sqrt{9^23^2}=6\sqrt{2}\); \(PQ=CP+CQ=12\sqrt{2}\); \(Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}\). Answer: A. P.S. Please post the new questions in separate threads. Attachment:
graph.php.png [ 17.5 KiB  Viewed 17431 times ]
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Re: The area of triangle POQ
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21 Oct 2010, 17:02
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
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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 \(\frac{OC}{OB}=\frac{OA}{AB}\) > \(OC=3\); \(CP=CQ=\sqrt{OP^2OC^2}=\sqrt{9^23^2}=6\sqrt{2}\); \(PQ=CP+CQ=12\sqrt{2}\); \(Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}\). 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
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23 Oct 2010, 04:35



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Re: Good Question ...Number system
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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
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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 \(\frac{OC}{OB}=\frac{OA}{AB}\) > \(OC=3\); \(CP=CQ=\sqrt{OP^2OC^2}=\sqrt{9^23^2}=6\sqrt{2}\); \(PQ=CP+CQ=12\sqrt{2}\); \(Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}\). 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
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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 \(\frac{OC}{OB}=\frac{OA}{AB}\) > \(OC=3\); \(CP=CQ=\sqrt{OP^2OC^2}=\sqrt{9^23^2}=6\sqrt{2}\); \(PQ=CP+CQ=12\sqrt{2}\); \(Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}\). 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: ifpandqaretwopointsontheline3x4y15suchthat103360.html#p805599
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Re: Good Question ...Number system
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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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Re: The area of triangle POQ
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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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Re: Good Question ...Number system
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23 Apr 2014, 06:54



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



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Re: Good Question ...Number system
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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 \(\frac{OC}{OB}=\frac{OA}{AB}\) > \(OC=3\); \(CP=CQ=\sqrt{OP^2OC^2}=\sqrt{9^23^2}=6\sqrt{2}\); \(PQ=CP+CQ=12\sqrt{2}\); \(Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}\). 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
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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 \(\frac{OC}{OB}=\frac{OA}{AB}\) > \(OC=3\); \(CP=CQ=\sqrt{OP^2OC^2}=\sqrt{9^23^2}=6\sqrt{2}\); \(PQ=CP+CQ=12\sqrt{2}\); \(Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}\). 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: ifarcpqraboveisasemicirclewhatisthelengthof144057.html#p1154669Hope it helps.
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Re: If P and Q are two points on the line 3x + 4y =15 such that
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27 Aug 2014, 07:11
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
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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 \(\frac{OC}{OB}=\frac{OA}{AB}\) > \(OC=3\); \(CP=CQ=\sqrt{OP^2OC^2}=\sqrt{9^23^2}=6\sqrt{2}\); \(PQ=CP+CQ=12\sqrt{2}\); \(Area_{OPQ}=\frac{base*height}{2}=\frac{PQ*OC}{2}=\frac{12\sqrt{2}*3}{2}=18\sqrt{2}\). 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
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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
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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) coordinates.



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Re: If P and Q are two points on the line 3x + 4y =15 such that
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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. Answer: Option A
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Re: If P and Q are two points on the line 3x + 4y =15 such that &nbs
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