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In the xy-coordinate plane above, what are the coordinates of point N

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Re: In the xy-coordinate plane above, what are the coordinates of point N [#permalink]
GMATinsight wrote:
Bunuel bb

I couldn't find the tag of OG2021 for this question for DS post which I could find for PS post. Please help

No tag yet. Please edit the image it's not visible.
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Re: In the xy-coordinate plane above, what are the coordinates of point N [#permalink]
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We require both statements to conclude answer.

Statement 1 is just telling about the triangle being an equilateral triangle. No cordinates or data given to determine point N. (not sufficient)

Statement 2 states Only cordinate of M. No data for N or P.(not sufficient)

Since MNP is an equilateral triangle. Distance between MO and PO will be equal which is 8 unit.
Area of triangle is 1/2*(base)*(height)=Area of equilateral triangle
1/2*MP*NO= √3/4*(side)^2
1/2*8*N0 = √3/4*(8)^2
NO=4√3
Since N is on y intercept.
X cordinate will be zero.
Y will be 4√3.

So IOC should be C

Posted from my mobile device
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Re: In the xy-coordinate plane above, what are the coordinates of point N [#permalink]
According to me there is something wrong in the question. It's not mentioned that "O" is passing through the origin so how can we be sure that point P has co-ordinates (4,0). What if it's co-ordinates are (2,0). Then the side=6 and not 8. And in the exam the figure is never drawn to scale basis. So we just can't rely on the image.

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Re: In the xy-coordinate plane above, what are the coordinates of point N [#permalink]
TaizKaran wrote:
According to me there is something wrong in the question. It's not mentioned that "O" is passing through the origin so how can we be sure that point P has co-ordinates (4,0). What if it's co-ordinates are (2,0). Then the side=6 and not 8. And in the exam the figure is never drawn to scale basis. So we just can't rely on the image.

Hi TaizKaran

There is nothing wrong in the question. This is official question

Since X and Y axis are meeting at Point O so it automatically becomes Origin.
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Re: In the xy-coordinate plane above, what are the coordinates of point N [#permalink]
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TaizKaran wrote:
According to me there is something wrong in the question. It's not mentioned that "O" is passing through the origin so how can we be sure that point P has co-ordinates (4,0). What if it's co-ordinates are (2,0). Then the side=6 and not 8. And in the exam the figure is never drawn to scale basis. So we just can't rely on the image.

Yes, you are correct that figures in DS question are not drawn to scale.

But the two statements combine tells you that P has co-ordinates (4,0).

It is an equilateral triangle, so the altitude will be in the middle of PM, so O has to be the center of MP.
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Re: In the xy-coordinate plane above, what are the coordinates of point N [#permalink]
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TaizKaran wrote:
According to me there is something wrong in the question. It's not mentioned that "O" is passing through the origin so how can we be sure that point P has co-ordinates (4,0). What if it's co-ordinates are (2,0). Then the side=6 and not 8. And in the exam the figure is never drawn to scale basis. So we just can't rely on the image.

You are given that it is the xy coordinate plane. y and x axis are drawn intersecting at O. Then O is definitely (0, 0). No questions about it.
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Re: In the xy-coordinate plane above, what are the coordinates of point N [#permalink]
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TaizKaran wrote:
According to me there is something wrong in the question. It's not mentioned that "O" is passing through the origin so how can we be sure that point P has co-ordinates (4,0). What if it's co-ordinates are (2,0). Then the side=6 and not 8. And in the exam the figure is never drawn to scale basis. So we just can't rely on the image.

Solution:

You are right that we cannot rely on the figure for certain information since figures are not necessarily drawn to scale; however, that doesn't mean that there's nothing about the figure that you can rely on. The figure clearly labels the perpendicular lines as the x-axis and the y-axis and since O is the point of intersections of the axes, it must be the origin. Even if the figure is not drawn to scale, the question cannot claim that the two perpendicular lines were just some arbitrary lines labeled "x" and "y"; those lines must be the x- and y-axes, respectively.

Some of the things you can assume of a figure are as follows: If a line is drawn as straight, then it is indeed straight (so that there are no hidden bends or curves). If a point is drawn on a line or a line segment, then that point is indeed on that line or line segment (so that the question cannot claim points M and P were "slightly above" the x-axis). If two lines or points are drawn as distinct, then they are indeed distinct (so that the question cannot claim points M and O were actually the same point). With these assumptions as well as the two statements, the answer is clearly C.
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Re: In the xy-coordinate plane above, what are the coordinates of point N [#permalink]
chetan2u

Wont we have to take Negative value of N... i.e. below x-axis....
Arent 2 equilateral triangles possible....?
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Re: In the xy-coordinate plane above, what are the coordinates of point N [#permalink]
zs2 wrote:
chetan2u

Wont we have to take Negative value of N... i.e. below x-axis....
Arent 2 equilateral triangles possible....?

No, we will not take the negative value of N, as it is shown on the positive side of x axis, and other 2 points are shown on x-axis.
had it been not given on sketch and it was given that 2 points are equidistant from the origin and lie on x-axis, then there would be possibility of N having negative x-coord.
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In the xy-coordinate plane above, what are the coordinates of point N [#permalink]
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Solution:

A new og 2021 question

St(1):- There is no information about coordinates of M,P. (Insufficient)

St(2):- We can shift the point N to any point on the Y axis with M remaining constant at (-4,0).

This would provide us with no unique value of N. (Insufficient)

Combining both we have

P has coordinates (4,0) as the perpendicular in an equilateral triangle(as given in St(1) bisects the base.

Given M is (-4,0) in St(2) =>P is (4,0).

Thus MP =4+4 = 8 units of length.

=> MN =8 units (equilateral triangle)

=>MO=4 units and hence ON can be computed ( Using Pythagorean theorem)

=> 1/2*8*N0 = √3/4*(8)^2

Since N is on y intercept, X coordinate will be 0 and hence N =(0, 4√3) (Sufficient) option (c)

Hope this helps
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Re: In the xy-coordinate plane above, what are the coordinates of point N [#permalink]
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Re: In the xy-coordinate plane above, what are the coordinates of point N [#permalink]
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