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Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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10 Jun 2015, 04:19
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Three of the four vertices of a rectangle in the xycoordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex? (A) (–4.5, 2.5) (B) ( –4, 5) (C) (6, –2 ) (D) (7, –3 ) (E) (10, 1) Kudos for a correct solution.
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Re: Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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10 Jun 2015, 07:59
Bunuel wrote: Three of the four vertices of a rectangle in the xycoordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?
(A) (–4.5, 2.5) (B) ( –4, 5) (C) (6, –2 ) (D) (7, –3 ) (E) (10, 1)
Kudos for a correct solution. ALTERNATELength of Diagonals are same in a Rectanglei.e. Distance Between ( –5, 1) and (8, 0) should be same as Distance between ( –4, 4) and 4th Vertex (x, y) i.e. \((01)^2+(8+5)^2 = (y4)^2+(x+4)^2\) i.e. \(1+169 = (y4)^2+(x+4)^2\) Check Options: 4th Vertex can only be in either 1st or 4th Quadrant so Option A and B are already out of sync Option (C)  (6, –2 ): (y4)^2+(x+4)^2 = (6)^2 + (10)^2 = 136 so INCORRECT(D) (7, –3 ): (y4)^2+(x+4)^2 = (7)^2 + (11)^2 = 49+121 = 170 so CORRECTAnswer: Option
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Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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10 Jun 2015, 07:51
Bunuel wrote: Three of the four vertices of a rectangle in the xycoordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?
(A) (–4.5, 2.5) (B) ( –4, 5) (C) (6, –2 ) (D) (7, –3 ) (E) (10, 1)
Kudos for a correct solution. ALTERNATIVESlope of Line joining points A( –5, 1) and B( –4, 4) = (41) / [(4)(5)] = 3/1 = m1 Slope of Line joining points B( –4, 4) and C( 8, 0) = (04) / [(8)(4)] = 4/12 = 1/3 = m2 This also suggests that 4th Point must be in Quadrant IV i.e Option C or D only can be truei.e. AB and BC lines are perpendicular as m1 * m2 = 1 Slope of Line joining points A(–5, 1) and D( x, y) = (y1) / (x+5) = m3 Slope of Line joining points C(8, 0) and D( x, y) = (y0) / (x8) = m4 Now m3 * m4 = 1 i.e. [(y1) / (x+5)] * [(y0) / (x8)] = 1 i.e \(y(y1) = (x+5)*(x8)\)Let's Check Option C (6, 2)LHS = (2)(21) = 6 RHS = (6+5)(68) = 22 INCORRECTLet's Check Option D (7, 3)LHS = (3)(31) = 12 RHS = (7+5)(78) = 12 CORRECTAnswer: Option
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Re: Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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13 Jun 2015, 06:43
Draw it on the paper, it takes a few seconds to "guess" the correct answer.
Then we could check the result with the formula for the Distance between two points.
Distance btw (5;1) and (4;4) is Sqrt(10)
then the correct point must be distant from (8;0) by Sqrt(10)



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Re: Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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13 Jun 2015, 08:15
Actually this is very easy to be solved if you draw on a paper. Answer D
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Re: Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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15 Jun 2015, 02:46
Bunuel wrote: Three of the four vertices of a rectangle in the xycoordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?
(A) (–4.5, 2.5) (B) ( –4, 5) (C) (6, –2 ) (D) (7, –3 ) (E) (10, 1)
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:Your GMAT scratchpad has a grid; use it to plot the diagram to scale. Attachment:
20150615_1342.png [ 13.23 KiB  Viewed 8818 times ]
“Eyeball” solution: The corner at ( –4, 4) looks like a right angle, so complete the rectangle with the dashed lines shown. The 4th point must be located approximately where the bigger dot is drawn. The closest answer choice is the point (7, –3 ). Alternatively, you could plot all of the answer choice points and see which one “works” with the three given points. Alternatively, we could solve by checking the slopes of the solid lines we drew between the given points to prove those lines are perpendicular, that is, to prove we have drawn two sides of the rectangle correctly. The slope of the long solid line = (4 0)/(4  8) = 1/3. The slope of the short solid line = (4  1)/(4  (5)) = 3. The product of these slopes is 1/3*3 = 1, proving that the lines are perpendicular. Compute the location of the 4th point, using the fact that the short sides have the same slope. The known short side connects the points ( –5, 1) and ( –4, 4). In other words, the bottom left corner is 1 to the left and 3 down from the top left corner. The unknown bottom right corner should therefore be 1 to the left and 3 down from the top right corner, or x = 8 – 1 = 7 and y = 0 – 3 = –3, corresponding to the point (7,–3). The correct answer is D.
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Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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28 Jun 2015, 12:19
Bunuel wrote: Bunuel wrote: Three of the four vertices of a rectangle in the xycoordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?
(A) (–4.5, 2.5) (B) ( –4, 5) (C) (6, –2 ) (D) (7, –3 ) (E) (10, 1)
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:Your GMAT scratchpad has a grid; use it to plot the diagram to scale. Attachment: 20150615_1342.png “Eyeball” solution: The corner at ( –4, 4) looks like a right angle, so complete the rectangle with the dashed lines shown. The 4th point must be located approximately where the bigger dot is drawn. The closest answer choice is the point (7, –3 ). Alternatively, you could plot all of the answer choice points and see which one “works” with the three given points. Alternatively, we could solve by checking the slopes of the solid lines we drew between the given points to prove those lines are perpendicular, that is, to prove we have drawn two sides of the rectangle correctly. The slope of the long solid line = (4 0)/(4  8) = 1/3. The slope of the short solid line = (4  1)/(4  (5)) = 3. The product of these slopes is 1/3*3 = 1, proving that the lines are perpendicular. Compute the location of the 4th point, using the fact that the short sides have the same slope. The known short side connects the points ( –5, 1) and ( –4, 4). In other words, the bottom left corner is 1 to the left and 3 down from the top left corner. The unknown bottom right corner should therefore be 1 to the left and 3 down from the top right corner, or x = 8 – 1 = 7 and y = 0 – 3 = –3, corresponding to the point (7,–3). The correct answer is D.Hi Bunuel, please explain why the rectangle cant be such that one side joins the points (5,1) and (8,0), and the opposite side joins the points (4,4) and the unknown? Thats what I did initially, but it led to a wrong answer. What (apparently obvious) clues did I miss? My approach was this: two parallel lines have equal slopes so slope for the line joining (5,1) and (8,0) = slope for the line joining the points (4,4) and the unknown. For the latter part (slope involving unknown point), x coordinate would be one unit more than (8,0), since the difference b/w x coordinates is 1 unit here: (5,1) and (4,4). Then solved for y (the only unknown variable left now). TIA!



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Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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28 Jun 2015, 21:59
mrish7 wrote: Bunuel wrote: Bunuel wrote: Three of the four vertices of a rectangle in the xycoordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?
(A) (–4.5, 2.5) (B) ( –4, 5) (C) (6, –2 ) (D) (7, –3 ) (E) (10, 1)
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:Your GMAT scratchpad has a grid; use it to plot the diagram to scale. Attachment: 20150615_1342.png “Eyeball” solution: The corner at ( –4, 4) looks like a right angle, so complete the rectangle with the dashed lines shown. The 4th point must be located approximately where the bigger dot is drawn. The closest answer choice is the point (7, –3 ). Alternatively, you could plot all of the answer choice points and see which one “works” with the three given points. Alternatively, we could solve by checking the slopes of the solid lines we drew between the given points to prove those lines are perpendicular, that is, to prove we have drawn two sides of the rectangle correctly. The slope of the long solid line = (4 0)/(4  8) = 1/3. The slope of the short solid line = (4  1)/(4  (5)) = 3. The product of these slopes is 1/3*3 = 1, proving that the lines are perpendicular. Compute the location of the 4th point, using the fact that the short sides have the same slope. The known short side connects the points ( –5, 1) and ( –4, 4). In other words, the bottom left corner is 1 to the left and 3 down from the top left corner. The unknown bottom right corner should therefore be 1 to the left and 3 down from the top right corner, or x = 8 – 1 = 7 and y = 0 – 3 = –3, corresponding to the point (7,–3). The correct answer is D.Hi Bunuel, please explain why the rectangle cant be such that one side joins the points (5,1) and (8,0), and the opposite side joins the points (4,4) and the unknown? Thats what I did initially, but it led to a wrong answer. What (apparently obvious) clues did I miss? My approach was this: two parallel lines have equal slopes so slope for the line joining (5,1) and (8,0) = slope for the line joining the points (4,4) and the unknown. For the latter part (slope involving unknown point), x coordinate would be one unit more than (8,0), since the difference b/w x coordinates is 1 unit here: (5,1) and (4,4). Then solved for y (the only unknown variable left now). TIA! Hi mrish7, Considering your argument, the slope of line joining points (5,1) and (8,0) = 1/13Now the third point (4,4) must be joined with (5, 1) or with (8, 0) to make the other side of rectangle and the new line must have slope of 13 because Product of slopes of two perpendicular lines = 1Case 1: (4,4)is joined with (5, 1), the slope of this line = 3/1
Case 2: (4,4)is joined with (8, 0) the slope of this line = 4/(12) = 1/3Both cases rejected. Therefore this assumption is INCORRECTI hope it helps!
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Re: Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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15 Oct 2016, 20:09
Isnt there is a formula to do the same? x coordinate (x1+x3x2) and y coordinate (y1+y2y3)??



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Re: Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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15 Oct 2016, 20:14
Just derived  The correct formula is x coordinate ( x1+x3x2) and y coordinate is (y1+y3y2).



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Re: Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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15 Oct 2016, 20:49
AmritaSarkar89 wrote: Just derived  The correct formula is x coordinate ( x1+x3x2) and y coordinate is (y1+y3y2). Although your previous score is already pretty decent but if you are considering taking GMAT again to improve it further then a small piece of advice is "Do away with formulas" as much as possible. A person getting the score as your profile shows can't be illogical and logical people should understand maths logically. It rewards them with greater score always In Coordinate geometry, ALWAYS DRAW THE FIGURE while solving questions... It makes your understanding very smooth and easy
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Re: Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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15 Oct 2016, 21:18
GMATinsight wrote: AmritaSarkar89 wrote: Just derived  The correct formula is x coordinate ( x1+x3x2) and y coordinate is (y1+y3y2). Although your previous score is already pretty decent but if you are considering taking GMAT again to improve it further then a small piece of advice is "Do away with formulas" as much as possible. A person getting the score as your profile shows can't be illogical and logical people should understand maths logically. It rewards them with greater score always In Coordinate geometry, ALWAYS DRAW THE FIGURE while solving questions... It makes your understanding very smooth and easy Thanks for your piece of advice. Yes I am planning on retaking to hit something close to 750. The problem is since our childhood we had been pushed into memorizing formulas and till date I am struggling to get away with that canopy on my thought process. Honestly yes, I am trying to logically breakdown every problem



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Re: Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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29 Apr 2018, 10:47
Three of the four vertices of a rectangle in the xycoordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?
(A) (–4.5, 2.5) (B) ( –4, 5) (C) (6, –2 ) (D) (7, –3 ) (E) (10, 1)
Since diagonals of rectangles are Equal ,we can calculate the distance between the diagonals and other two points of vertex must also satisfy that distance . Points ( –5, 1), and (8, 0) have a distance of sq root of 170 and similarly point ( –4, 4) third vertex and fourth vertex points mentioned in the option must satisfy this distance which is option D(7,3).



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Re: Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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13 May 2018, 09:34
As it’s a rectangle, opposite sides are equal and parallel. So the slope of the opposites is going to be same m1=m2= 04/8+4 =1/3 As we have the slope and (5,1) eq of line obtained: 3y+2x=2. As it’s already clear that the point will lying in the fourth quadrant, so eliminate options and substitue it on the eq of the line.



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Re: Three of the four vertices of a rectangle in the xycoordinate plane [#permalink]
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13 May 2018, 14:57
Without solving, we have y = 1, 4, 0. 14 = 3; hence, we need 3 as the difference in y which implies that the last y must be 3 or 3. Only D makes this true so D.
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Re: Three of the four vertices of a rectangle in the xycoordinate plane
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