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HoneyLemon
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ArjunKumar25
I really think the answer is wrong. I tried plotting this graph on desmos and its not possible for it be a square with these coordinates


YESS.....ans should be (-7,-4) or (-2,-4) or the area of the sq. should be 16 not 25.
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This is a strange question... an evil one maybe? I also assumed that the square was going to be located straight on and there were 4 options.
Instead, it seems the question assumes that the distance between (-2, 1) is more like a center of a circle and then another vertex can be 5 units away in any direction.

I have drawn 2 squares - a blue one (correct answer)
And a yellow one which is the exact same square (I copied it and pasted it but then removed the tilt without changing the size), so it is possible but it is not pretty.
P.S. Thank you ArjunKumar25 for the useful tool. Made it in desmos

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HoneyLemon
A square with area 25 has one vertex on point (-2, 1) in the coordinate plane. From the table below, select the x- and y-coordinates that could correspond to another vertex of this same square.


There could be two ways the square could be made.

1) Sides parallel to x and y axis
The coordinates then should be such that one of the two remains constant and other increases/decreases by 5. For example (-2+5,1) or (3,1), (-7,1), (-2,6), (-2,-4). It could also be diagonally opposite vertex but then it will not be integers. Nothing in option.

2) Sides slanting to x and y axis
Now, the side will become a hypotenuse. So hypotenuse is 5, as the side is 5.
The triangle made by this hypotenuse could have sides in ratio 3:4:5.
Now x could be 3 or 4. => If it were 4, -2-x=4 or x-2=4 would require x to be even. But the options are all odd.
So x should become 3 using one of the options. -2-x=5 or x=-7.
Also y can become 4 from given options. 1-y=4 or y=-3.

See the attached figure for reference.
Attachments

0DA5391D-695F-4A86-8F2E-99CFFAA5B60F.png
0DA5391D-695F-4A86-8F2E-99CFFAA5B60F.png [ 105.96 KiB | Viewed 3665 times ]

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evil is the right word.. how do we calculate the co-ordinates when the square is of this form?
bb
This is a strange question... an evil one maybe? I also assumed that the square was going to be located straight on and there were 4 options.
Instead, it seems the question assumes that the distance between (-2, 1) is more like a center of a circle and then another vertex can be 5 units away in any direction.

I have drawn 2 squares - a blue one (correct answer)
And a yellow one which is the exact same square (I copied it and pasted it but then removed the tilt without changing the size), so it is possible but it is not pretty.
P.S. Thank you ArjunKumar25 for the useful tool. Made it in desmos

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Hello!
I solved this question using the distance formula.
We know that the side of the square is 5 units, since side = sqrt (area) = sqrt (25) = 5

Let the coordinates of the vertex we want to find be (x,y). We know the already existing coordinates are (-2, 1)
Now, Using the distance formula,
sqrt( (x - (-2))^2 + (y-1)^2 ) = 5
Squaring both side,
(x+2)^2 + (y-1)^2 = 25

Now among the options, we can completely discard -7 and -9 since the LHS will become more than 25 in both the cases.
We are left with -1, -3 and -5.
We also know that 25 = 3^2 + 4^2
So, x+2 is equal to either 3 or -3, or 4 or -4
From the options, only x+2 = -3 is the feasible option
Hence, x = -5
Similarly, y-1 is equal to either 4 or -4
From the options, we see -3 is fulfiling this criteria.
Hence, y = -3

Hope this helps
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