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Re: Coordinate Plane - Find the point with distance [#permalink]

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10 Sep 2011, 08:39

hoaivinh wrote:

Hi folks,

This is a rather hard question on coordinate plane and I need your help. "What are the coordinates for the point on line AB that is 3 times as far as from B, and that is between points A and B, knowing that A(-5,6) and B(-2,0)"

At first, I worked out the x-axis distance from A to B is 3 (-5+2) and y-axis distance from A to B is 6 (6-0). I divide 4 segments, x for per segments, then I have 4x. 3x from A and x from B. Then I get stuck there. But I do not understand why the explanation in the answer is that "3x=4, x=0.75. The point is located on x-axis -2.75". I really do not understand.

Thank you.

The x-coordinate of the point is x=0.75 away from -2, and thus it's located on -2.75
_________________

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Re: Coordinate Plane - Find the point with distance [#permalink]

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10 Sep 2011, 17:08

1

This post was BOOKMARKED

Silver89 wrote:

hoaivinh wrote:

Hi folks,

This is a rather hard question on coordinate plane and I need your help. "What are the coordinates for the point on line AB that is 3 times as far as from B, and that is between points A and B, knowing that A(-5,6) and B(-2,0)"

At first, I worked out the x-axis distance from A to B is 3 (-5+2) and y-axis distance from A to B is 6 (6-0). I divide 4 segments, x for per segments, then I have 4x. 3x from A and x from B. Then I get stuck there. But I do not understand why the explanation in the answer is that "3x=4, x=0.75. The point is located on x-axis -2.75". I really do not understand.

Thank you.

The x-coordinate of the point is x=0.75 away from -2, and thus it's located on -2.75

Thank you Silver89. But how is 3x=4? That's the thing I got stuck.

Note the following points about posting. 1) If you want to attach a figure with the post, just copy and paste the figure in MS Paint then save the file as png image. Then simply type the question stem in the post text box and attach the saved png image to the post 2) Post the question in relevant forum. e.g. This question should be posted in GMAT quantitative - Problem solving subforum, and should not be in General GMAT questions. 3) Post the question along with OA.

What are the coordinates for the point on Line AB [#permalink]

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28 Nov 2014, 00:47

Hi experts, I found this approach in a flashcard to this question: a) The total x-axis distance from one point to the other is -3 (-5 - -2) b) The total y-axis distance from one point to other is 6 (6 – 0) c) Considering that the distance between A and B is 4x, and we need a point which is 3x distant from A and x distance from B. d) 4x = -3 --> x = -0.75. (Have corrected the typo here seeing a post above) The point is located on x-axis at -2.75 e) 4y = 6 --> y = 1.5. The point is located on y-axis at 1.5 f) The answer is (-2.75, 1.5)

Kindly help me understand how the 0.75 becomes -2.75 and how the 1.5 remains as 1.5. _________________

Cheers!!

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Re: What are the coordinates for the point on Line AB [#permalink]

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22 Jun 2016, 02:47

Can somebody please tell me where I am going wrong. Taking ABC with A being ( -5,6) , B being (-2,0) and C being (-5,0) . Length of AC is 6 cm. Length of CB is 3 cm . Using Pytho theorem we can get that -

6^2 + 3^2 = AB^2 36 + 9 = 45 = AB ^ 2

So, AB = 4X = X = sqrt(45) /4

Now using similiarity , we can find that the x and y co-ordinates of the smaller triangle. Mark the small triangle as BED .

ED / 6 = [ sqrt(45)/4 ] / sqrt(45) = 1/4 ED or the Y co-ordinate for the point D will be = 6/4 = 1.5

EB / 3 = 1/4 EB or X co-ordinate for the point D will be = 3/4 = 0.75

I know I am wrong somewhere ! Can somebody please point it out to clear my concept. Thanks in advance

It is critical to know the internal section formula in order to solve this question. Basically, this formula allows us to solve for a point on a line such as line AB , let's say point "P" , provided we know the ratio of the distance between that point and the other two end points on that line ( the end points are the coordinates given in the problem ) which is this is scenario AP to BP. Now, there's two ways of understanding the formula- what the question is referring to when it states a point that is "three times as far from A as from B" is that if you split line AB into two parts then the length between A and the hypothetical (imaginary point in question) P is 3 times greater than P to point B. Algebraically, we can say this ratio is 3x to x, or really just more simply 3 to 1. But as we look at the diagram how do we really understand each component of the internal section formula? That is to say how do we appropriately plug in 3 and 1 to m and n. Should 3 be m or should 3 be n? If we reverse engineer the formula

[{m(x2) + n(x1) / m + n} , {m(y2) + n (y1)/ m +n}] - the sum of our ratio, m and n, equals the denominator of the fraction.

But wait this is also the same thing as

m/ m + n (x2, y2) + n/m +n (x1,y2)

The key to properly assigning each x coordinate to either m or n respectively depends on where you partition the line- this is why a triangle helps. So if we take coordinate A for example such as in the both of my diagrams and the problem, we would align the coordinate of A with the number that corresponds to ratio of the line that is across the partition.

Here is an updated diagram- hopefully the color coding helps make it more clear how to correctly plug in each set of coordinates with correct ratio- here we ca see m is 2 and therefore it must be matched up with the coordinate pair as per the internal section formula

[2/2+ 3 (7,9)] , [ 3/2 +3 (5,3)] = m/ m + n (7,9) , n/ m + n (5,3) = P(x,y) - here p is the imaginary point between A and B

OR

mx2x1/ m + n , my2y1/ m + n = P(x,y) - mathematically this is the same as the formula but just compressed- here's why

2 (7) / (2+3) + 3 (5)/ 2 +3) = 14/ (2 +3) + 15/ (2+3) is the same thing as 14 + 15/ 2+ 3 - and when you add these you get the x coordinate so instead of multiplying it as through the first formula the second formula just compresses everything- but for the sake of understanding the concept the first formula may be better to start with.

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Re: What are the coordinates for the point on Line AB [#permalink]

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16 Jun 2017, 23:13

gmatprep2o16 wrote:

Can somebody please tell me where I am going wrong. Taking ABC with A being ( -5,6) , B being (-2,0) and C being (-5,0) . Length of AC is 6 cm. Length of CB is 3 cm . Using Pytho theorem we can get that -

6^2 + 3^2 = AB^2 36 + 9 = 45 = AB ^ 2

So, AB = 4X = X = sqrt(45) /4

Now using similiarity , we can find that the x and y co-ordinates of the smaller triangle. Mark the small triangle as BED .

ED / 6 = [ sqrt(45)/4 ] / sqrt(45) = 1/4 ED or the Y co-ordinate for the point D will be = 6/4 = 1.5

EB / 3 = 1/4 EB or X co-ordinate for the point D will be = 3/4 = 0.75

I know I am wrong somewhere ! Can somebody please point it out to clear my concept. Thanks in advance

WHy are you trying to solve this way ..

Something is not clear in your solution. You are just dividing the distance between AB and taking it as coordinate...

Try to use internal section formula.. it is much easier to solve and will save you a lots of time. After all getting a good score in GMAT is all about time. So try to find shorter methods to solve question and move towards next question ASAP. Otherwise getting a 50 or 51 in QA will be difficult.
_________________

What are the coordinates for the point on Line AB [#permalink]

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11 Dec 2017, 10:41

Hi! Where this problem comes from? Not official GMAT? Do real GMAT problems like that appear on the exam so that I need to memorize this internal section formula (which by the way is never included in any theoretical basis for GMAT and even unofficial flashcards (this problem is solved differently not via formula there?)? I also found it is easier to resolve this question by simple dividing by 4 of x and y distance and adding to existing point... I do not see any need to know this internal section formula here at all..