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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
Bunuel wrote:
A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) respectively. What is the area of the square?

A. 17
B. 21
C. 27
D. 31
E. 34


draw fig we see tht given points depict the digaonal so distance; √34
and s√2= √34
s= √17
so area ; 17
IMO A
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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
The sides are not parallel to X and y axis. Can be concluded by plotting the points on the graph.
Hence one side of the square = Distance between the two points . √(5^2 + 3^2) = √34
Area of the square = side^2 = 34

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A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
nick1816 wrote:
\(Area= AB^2\)
\(AB^2= {4-(-1)}^2+{2-(-1)}^2\)
\(AB^2= 5^2+3^2\)
\(Area=AB^2=34\)
Bunuel wrote:
A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) respectively. What is the area of the square?

A. 17
B. 21
C. 27
D. 31
E. 34


nick1816 Sithara286
how can points vertices A and B at (–1, –1) and (4, 2) be on same line ? wont he quadrants be different
for (-1,-1) to have a vertice on same plane the cordiante has to be -ve isnt it ? i mean -1,-1 will fall in 3rd quadrant so the point to be on same line as -1,-1 the other point has to be either 4th or 2nd quadrant..
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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
You assumed that the sides of squares are parallel to co-ordinate axis; it's not mentioned in the question.

Archit3110 wrote:
nick1816 wrote:
\(Area= AB^2\)
\(AB^2= {4-(-1)}^2+{2-(-1)}^2\)
\(AB^2= 5^2+3^2\)
\(Area=AB^2=34\)
Bunuel wrote:
A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) respectively. What is the area of the square?

A. 17
B. 21
C. 27
D. 31
E. 34


nick1816 Sithara286
how can points vertices A and B at (–1, –1) and (4, 2) be on same line ? wont he quadrants be different
for (-1,-1) to have a vertice on same plane the cordiante has to be -ve isnt it ? i mean -1,-1 will fall in 3rd quadrant so the point to be on same line as -1,-1 the other point has to be either 4th or 2nd quadrant..
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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
nick1816 in that case whats the point of giving coordinate points? this question seems to be ambiguous

[quote="nick1816"]You assumed that the sides of squares are parallel to co-ordinate axis; it's not mentioned in the question.
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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
This question is not ambiguous at all. Co-ordinate points are given so that you can apply the distance formula to find the length of the side of square. Never assume anything in GMAT questions unless it's given.
Archit3110 wrote:
nick1816 in that case whats the point of giving coordinate points? this question seems to be ambiguous

nick1816 wrote:
You assumed that the sides of squares are parallel to co-ordinate axis; it's not mentioned in the question.
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A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
Bunuel wrote:
A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) respectively. What is the area of the square?

A. 17
B. 21
C. 27
D. 31
E. 34

Originally posted by firas92 on 11 Jun 2019, 10:04.
Last edited by firas92 on 11 Jun 2019, 12:06, edited 1 time in total.
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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
nick1816 ; but isnt distance formula used for two points which are plotted on a cartesian plane ? if so then we need to plot those points and calculate distance accordingly.. which in this case would be distance of diagonal not side...



nick1816 wrote:
This question is not ambiguous at all. Co-ordinate points are given so that you can apply the distance formula to find the length of the side of square. Never assume anything in GMAT questions unless it's given.
Archit3110 wrote:
nick1816 in that case whats the point of giving coordinate points? this question seems to be ambiguous

nick1816 wrote:
You assumed that the sides of squares are parallel to co-ordinate axis; it's not mentioned in the question.
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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
firas92 wrote:
If the coordinates (-1,-1) and (4,2) were diagonally opposite,
they won't form a square anyway

Therefore they must be the coordinates for a side.


Answer is (E)

Posted from my mobile device


firas92 could you please elaborate the highlighted part on why so ?
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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
Bunuel wrote:
A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) respectively. What is the area of the square?

A. 17
B. 21
C. 27
D. 31
E. 34


In GMAT question whenever a quadrilateral ABCD is given, we only assume that
—> Adjacent sides are AB, BC, CD and DA
—> Diagonals are AC and BD.

Given AB has to be side of square only.
We can’t assume AB as diagonal

So, option E seems perfectly fine.

Posted from my mobile device
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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
Dillesh4096 ,
I am not somehow convinced with the reasoning posted by all who have given option E as answer..
my points of doubt ;
1.considering it to be a cartesian plane how can points (–1, –1) and (4, 2) fall under same line and be side of a square ' graphically how is it possible'?
2. if you say that points do not fall under cartesian plane then how is the logic of distance formula valid ?
3. i agree that adjacent sides are mentioned in most questions as a,b,c,d but it cannot always be true as well..

irrespective of what might be the correct option of this question be , is there any logical reason why option e stands correct over a ?


Dillesh4096 wrote:
Bunuel wrote:
A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) respectively. What is the area of the square?

A. 17
B. 21
C. 27
D. 31
E. 34


In GMAT question whenever a quadrilateral ABCD is given, we only assume that
—> Adjacent sides are AB, BC, CD and DA
—> Diagonals are AC and BD.

Given AB has to be side of square only.
We can’t assume AB as diagonal

So, option E seems perfectly fine.

Posted from my mobile device
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A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
1
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Archit3110 wrote:
Dillesh4096 ,
I am not somehow convinced with the reasoning posted by all who have given option E as answer..
my points of doubt ;
1.considering it to be a cartesian plane how can points (–1, –1) and (4, 2) fall under same line and be side of a square ' graphically how is it possible'?
2. if you say that points do not fall under cartesian plane then how is the logic of distance formula valid ?
3. i agree that adjacent sides are mentioned in most questions as a,b,c,d but it cannot always be true as well..

irrespective of what might be the correct option of this question be , is there any logical reason why option e stands correct over a ?


Dillesh4096 wrote:
Bunuel wrote:
A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) respectively. What is the area of the square?

A. 17
B. 21
C. 27
D. 31
E. 34


In GMAT question whenever a quadrilateral ABCD is given, we only assume that
—> Adjacent sides are AB, BC, CD and DA
—> Diagonals are AC and BD.

Given AB has to be side of square only.
We can’t assume AB as diagonal

So, option E seems perfectly fine.

Posted from my mobile device


1.considering it to be a cartesian plane how can points (–1, –1) and (4, 2) fall under same line and be side of a square ' graphically how is it possible'?

Let’s take vertices C as (m, n).
Slope of AB = 3/5.
So slope of BC = -5/3
—> (n - 2)/(m - 4) = -5/3
—> 3n - 6 = -5m + 20
—> 3n + 5m = 26

Distance between AB = BC = sqrt(34)
—> BC^2 = 34
—> (m - 4)^2 + (n - 2)^2 = 34
—> (26/5- 3n/5 - 4)^2 + (n - 2)^2 = 34
—> (6 - 3n)^2 + 25(n - 2)^2 = 25*34
—> 9(n - 2)^2 + 25(n - 2)^2 = 25*34
—> (n - 2)^2 = 25
—> n - 2 = 5 or -5
—> n = 7 or -3
If n = 7, m = 1.
If n = -3, m = 7.
So C = (1, 7) or (7, -3)

Distance between A (-1, -1) and C(1, 7) or C (7, -3) = sqrt(68) which satisfies diagonal for option E.

Similarly you can find vertices D also.

Point is square need not be parallel to x and y axis in Cartesian system.


2. if you say that points do not fall under cartesian plane then how is the logic of distance formula valid ?

They do as per above C(1, 7) or (7, -3)

3. i agree that adjacent sides are mentioned in most questions as a,b,c,d but it cannot always be true as well..

Can you pls show any official question which doesn’t follow that order. Would love to see that as I haven’t seen any question till now.

irrespective of what might be the correct option of this question be , is there any logical reason why option e stands correct over a ?

For the same reason that I haven’t come across any question that GMAT said ABCD is square and considered AB as a diagonal rather than side.

Originally posted by CareerGeek on 11 Jun 2019, 11:30.
Last edited by CareerGeek on 11 Jun 2019, 11:31, edited 1 time in total.
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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
1
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Bhai just to clear your doubt, itna hardwork kiya hai LOL!!! I'm attaching the diagram of the square in Co-ordinate plane.
Archit3110 wrote:
Dillesh4096 ,
I am not somehow convinced with the reasoning posted by all who have given option E as answer..
my points of doubt ;
1.considering it to be a cartesian plane how can points (–1, –1) and (4, 2) fall under same line and be side of a square ' graphically how is it possible'?
2. if you say that points do not fall under cartesian plane then how is the logic of distance formula valid ?
3. i agree that adjacent sides are mentioned in most questions as a,b,c,d but it cannot always be true as well..

irrespective of what might be the correct option of this question be , is there any logical reason why option e stands correct over a ?


Dillesh4096 wrote:
Bunuel wrote:
A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) respectively. What is the area of the square?

A. 17
B. 21
C. 27
D. 31
E. 34


In GMAT question whenever a quadrilateral ABCD is given, we only assume that
—> Adjacent sides are AB, BC, CD and DA
—> Diagonals are AC and BD.

Given AB has to be side of square only.
We can’t assume AB as diagonal

So, option E seems perfectly fine.

Posted from my mobile device

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Capture1.PNG [ 15.44 KiB | Viewed 3387 times ]

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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
Archit3110 wrote:
firas92 wrote:
If the coordinates (-1,-1) and (4,2) were diagonally opposite,
they won't form a square anyway

Therefore they must be the coordinates for a side.


Answer is (E)

Posted from my mobile device


firas92 could you please elaborate the highlighted part on why so ?


I'm sorry that reasoning might be wrong. In fact here's a square with given coordinates as a diagonal. Im more confused now :lol:
Attachments

Untitled.png
Untitled.png [ 10.47 KiB | Viewed 3345 times ]

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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
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But the square you drew is ACBD, not ABCD that is given in the question.
firas92 wrote:
Archit3110 wrote:
firas92 wrote:
If the coordinates (-1,-1) and (4,2) were diagonally opposite,
they won't form a square anyway

Therefore they must be the coordinates for a side.


Answer is (E)

Posted from my mobile device


firas92 could you please elaborate the highlighted part on why so ?


I'm sorry that reasoning might be wrong. In fact here's a square with given coordinates as a diagonal. Im more confused now :lol:
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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
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firas92 wrote:
Archit3110 wrote:
firas92 wrote:
If the coordinates (-1,-1) and (4,2) were diagonally opposite,
they won't form a square anyway

Therefore they must be the coordinates for a side.


Answer is (E)

Posted from my mobile device


firas92 could you please elaborate the highlighted part on why so ?


I'm sorry that reasoning might be wrong. In fact here's a square with given coordinates as a diagonal. Im more confused now :lol:


firas92

So, if you consider AB as diagonal you can definitely find a square with AB as diagonal. Problem is GMAT doesn’t consider AB as diagonal when they mention ABCD is a square/quadrilateral.

Only if they mention ACBD is a square, AB would be diagonal.
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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
nick1816 Dillesh4096


Understood.

Thanks for the clarification. I see how the wording matters now!

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Re: A square ABCD has two of its vertices A and B at (–1, –1) and (4, 2) [#permalink]
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