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Director
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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10 Aug 2012, 21:56
VeritasPrepKarishma wrote: rajathpanta wrote: bscharm wrote: Theoretically couldn't we know that (2,1) and (4,1) make up the diameter of the circle? Because both points are "on the circle" as the question states, no two points on that axis (where y=1) can be closer or further away, hence it must be the widest part of the circle and the diameter. From there we can solve the midpoint of those two coordinates (making point G unnecessary) so...
(2+4)/2 = 1 for x
(1+1)/2 = 1 for y
so (1,1) must be the center of that line as well as the circle. Hi, Even I had the same question. Now I am a little confused. Can anyone explain this concept please!! It is certainly correct that both points lie on the circle and no points on the circle on y=1 can be closer or farther away. But what says that these points make a diameter and not just a chord smaller than the diameter? If you do find their mid point, you need to check whether the distance from the third point is equal too. As I said in my first post, you can very easily solve the question by just plotting the points. You can see that all points are equidistant from (1, 1) (Check out a figure I made in a previous post) As for the distance formula, it is a great little tool for any scenario so it's good to know. Plotting the points is almost a must in order to get a feeling what the triangle looks like. Why not check first that FG is perpendicular to GH? Just an easy test on the slopes. Than immediately follows that FG is perpendicular to GH, so FH must be diameter. Please, refer to my previous posts above.
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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10 Aug 2012, 22:14
EvaJager wrote: Plotting the points is almost a must in order to get a feeling what the triangle looks like.
Why not check first that FG is perpendicular to GH? Just an easy test on the slopes. Than immediately follows that FG is perpendicular to GH, so FH must be diameter.
Please, refer to my previous posts above. Because there is no test needed once you plot it. You can SEE that every point is 3 steps away from (1, 1). Anyway, everybody approaches questions differently and our opinions on the best method are bound to be different. So the best strategy is to consider various opinions and incorporate what you feel works best for you. It needn't be the same for everyone.
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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10 Aug 2012, 22:35
VeritasPrepKarishma wrote: EvaJager wrote: Plotting the points is almost a must in order to get a feeling what the triangle looks like.
Why not check first that FG is perpendicular to GH? Just an easy test on the slopes. Than immediately follows that FG is perpendicular to GH, so FH must be diameter.
Please, refer to my previous posts above. Because there is no test needed once you plot it. You can SEE that every point is 3 steps away from (1, 1). Anyway, everybody approaches questions differently and our opinions on the best method are bound to be different. So the best strategy is to consider various opinions and incorporate what you feel works best for you. It needn't be the same for everyone. Agree that not everybody should think the same. It would be a pity to ignore a fast method. Once you plot the points, in this case, it is really obvious which point is the center of the circumscribed circle. If none of the sides of the triangle would have been parallel to the axis, then checking distances is not the fastest method. If the triangle is a right triangle, which is mostly the case on the GMAT, then testing the slopes is the faster and easiest way to determine where is the right angle, which one is the hypotenuse,...Or just look for the right triangle. No arguing about the property of the center of the circumscribed circle (being equidistant from the vertices of the triangle), but it isn't the fastest way to find it, definitely not on a GMAT question. And it is worth remembering the property of the right triangle (hypotenuse diameter in the circumscribed circle) and it is worth checking the type of the triangle before we launch into lengthy computations.
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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11 Aug 2012, 02:22
VeritasPrepKarishma wrote: EvaJager wrote: Plotting the points is almost a must in order to get a feeling what the triangle looks like.
Why not check first that FG is perpendicular to GH? Just an easy test on the slopes. Than immediately follows that FG is perpendicular to GH, so FH must be diameter.
Please, refer to my previous posts above. Because there is no test needed once you plot it. You can SEE that every point is 3 steps away from (1, 1). Anyway, everybody approaches questions differently and our opinions on the best method are bound to be different. So the best strategy is to consider various opinions and incorporate what you feel works best for you. It needn't be the same for everyone. Can you easily SEE here where is the center of the circumscribed circle? circlepassesthroughpoints1225and42105.htmlWould you also go into distance calculation in this case? Isn't Bunuel's solution the fastest?
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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11 Aug 2012, 06:17
EvaJager wrote: Can you easily SEE here where is the center of the circumscribed circle? circlepassesthroughpoints1225and42105.htmlWould you also go into distance calculation in this case? Isn't Bunuel's solution the fastest? Kindly refrain from getting argumentative. I don't recall suggesting that anyone else's method is not good or appropriate. Every question is different and the strategy one uses for different questions is different. What I choose to do in the other question is immaterial. Let me say it once again and for the last time  I give solutions which in my opinion work the best for that particular question. I don't think I am forcing you to consider my point of view.
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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11 Aug 2012, 07:41
VeritasPrepKarishma wrote: EvaJager wrote: Can you easily SEE here where is the center of the circumscribed circle? circlepassesthroughpoints1225and42105.htmlWould you also go into distance calculation in this case? Isn't Bunuel's solution the fastest? Kindly refrain from getting argumentative. I don't recall suggesting that anyone else's method is not good or appropriate. Every question is different and the strategy one uses for different questions is different. What I choose to do in the other question is immaterial. Let me say it once again and for the last time  I give solutions which in my opinion work the best for that particular question. I don't think I am forcing you to consider my point of view. Every triangle can be inscribed in a circle. Right triangles are very special, as their circumscribed circles can be very easily determined  specifically, having the diameter the hypotenuse. After Pythagoras's theorem, this is the second most important property one should remember about right triangles.When asked whether a certain side of a triangle can be the diameter of the circumscribed circle, the answer should be that this definitely can be tested, by checking whether the opposite angle is a right angle or not (testing the slopes). Ignoring this fact doesn't benefit those who want to learn from our posts. They will probably continue to plug an chug, as being fixated on the sole property of equidistance property. I will certainly continue to post information that would benefit those preparing for the test.
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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26 Jan 2015, 10:49
Find the slopes of FG and GH I.e 1 and 1 respectively
so the above slopes are negative reciprocal of the points , thus they are perpendicular and if two perpendicular lines are inscribed in a circle the adjoining line will make it a right angle triangle and thus it will be the diameter of the circle.
so diameter is FH. find the midpoints of these two points ,Ie (1,1) hence the shortest method.



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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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29 Apr 2016, 13:45
Hi Karishma I think your way is more simple and time saving but I gave glance to the three points on the circle and noticed that (1,4),(4,1) are mirror view which needs X and Y to be equal to produce equal distance between both points which are radius, using your formula. Is it right in geometry or coordinate geometry to use observation to help solve the problem especially that such question test the geometrical vision i mean intuitive view.



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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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29 Apr 2016, 13:45
Hi Karishma I think your way is more simple and time saving but I gave glance to the three points on the circle and noticed that (1,4),(4,1) are mirror view which needs X and Y to be equal to produce equal distance between both points which are radius, using your formula. Is it right in geometry or coordinate geometry to use observation to help solve the problem especially that such question test the geometrical vision i mean intuitive view.



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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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08 May 2016, 16:38
Sorry, a very fundamental question. In the exam how can we solve such questions by 'drawing it'?. I know we are given erasable graph page, but will it be quick enough to solve it by drawing on it? VeritasPrepKarishma wrote: gettinit wrote: In the coordinate plane, the points F (2,1), G (1,4), and H (4,1) lie on a circle with center P. What are the coordinates of point P ? (A) (0,0) (B) (1,1) (C) (1,2) (D) (1,2) (E) (2.5, (2.5) Remember that center of the circle is equidistant from any point ont he circle. This distance is of course the radius of the circle. The fastest method here is to use the options to see which point is equidistant from F (2,1), G (1,4) and H (4,1). Formula for distance between two points is given by \(\sqrt{(x1  x2)^2 + (y1  y2)^2}\) I see that (0, 0) will not be equidistant from the given 3 points. But distance of (1, 1) from (2, 1) is \(\sqrt{(2  1)^2 + (1  1)^2}\) = 3 Distance of (1, 1) from (1, 4) is \(\sqrt{(1  1)^2 + (1  4)^2}\) = 3 Distance of (1, 1) from (4, 1) is \(\sqrt{(4  1)^2 + (1  1)^2}\) = 3 The coordinates of the center of the circle, P, must be (1, 1). (What I actually did was plotted the points on the xy axis and then saw that (1, 1) will be equidistant from all three points because (2, 1) lies directly 3 steps to its left, (1, 4) lies directly 3 steps above it and (4, 1) lies directly 3 steps to its right.



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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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08 May 2016, 22:37
hatemnag wrote: Hi Karishma I think your way is more simple and time saving but I gave glance to the three points on the circle and noticed that (1,4),(4,1) are mirror view which needs X and Y to be equal to produce equal distance between both points which are radius, using your formula. Is it right in geometry or coordinate geometry to use observation to help solve the problem especially that such question test the geometrical vision i mean intuitive view. Absolutely! It is not "just a feeling". It is logical. Since the two points (with x and y coordinates reversed) are mirror image using y = x as axis, they will be equidistant from a point which has same x and y coordinates. Now, see that (1, 4) is exactly 3 units away from (1, 1) vertically. They have the same x coordinate and (1, 4) is located 3 units above (1, 1). (2, 1) is exactly 3 units away from (1, 1) horizontally to the left. There is NO approximation. Both points have the same y coordinate. We could very well have observed this without making a diagram, just by looking at the coordinates given but that is much harder. A diagram makes it easier to see the relative placement and then infer.
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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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09 Jan 2017, 18:52
I've learnt to make sure to spot any similarities in the Y or X coordinates of the 3 points given. If there are, then the problem becomes a lot easier. This is true for triangle problems too with coordinate geometry.



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Re: In the coordinate plane, the points F (2,1), G (1,4), and H
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23 Sep 2018, 00:32
General Circle Equation  (xa)^2 +(yb)^2=r^2 Where a,b = center of the circle r= radius. The radius should be same from all given points from the center.
P1. (2,1) (2a)^2+(1b)^2=r^2 p2. (1,4) (1a)^2+(4b)^2=r^2. p3. (4,1) (4a)^2 +(1b)^2=r^2.
By elimination process : (1,1) radius remains the same for all the points on the circle.




Re: In the coordinate plane, the points F (2,1), G (1,4), and H &nbs
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