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The vertices of square S have coordinates
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06 May 2017, 11:35
06 May 2017, 11:35
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Question Stats:
51% (01:29) correct
48% (01:35) wrong based on 33 sessions
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The vertices of square S have coordinates (-1,-2), (-1,1), (2,1), and (2,-2) respectively. What are the coordinates of the point where the diagonals of S intersect?
a) (1/2, 1/2)
b) (1/2, -1/2)
c) (3/2, 1/2)
d) (3/2, -1/2)
e) (underoot3/2, 1/2)
a) (1/2, 1/2)
b) (1/2, -1/2)
c) (3/2, 1/2)
d) (3/2, -1/2)
e) (underoot3/2, 1/2)
Re: The vertices of square S have coordinates
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07 May 2017, 02:05
07 May 2017, 02:05
Expert Reply
HarveyKlaus wrote:
The vertices of square S have coordinates (-1,-2), (-1,1), (2,1), and (2,-2) respectively. What are the coordinates of the point where the diagonals of S intersect?
a) (1/2, 1/2)
b) (1/2, -1/2)
c) (3/2, 1/2)
d) (3/2, -1/2)
e) (underoot3/2, 1/2)
a) (1/2, 1/2)
b) (1/2, -1/2)
c) (3/2, 1/2)
d) (3/2, -1/2)
e) (underoot3/2, 1/2)
The easiest way to go about this question is to sketch it and identify which are the diagonally opposite vertices. However there is a shorter way to accomplish the same.
We know that diagonals of a square bisect each other. So in a square ABCD, the coordinates of the point where the diagonals of ABCD intersect.
\(\frac{A + C}{2} = \frac{B + D}{2}\)
or coordinates of the point where the diagonal intersects can be written as = \(\frac{A+B+C+D}{4}\).
So \(x= \frac{-1 -1 +2 +2}{4}=\frac{1}{2}\)
\(y= \frac{-2 +1 +1 -2}{4}=\frac{-1}{2}\).
Hence option B is correct!
Re: The vertices of square S have coordinates
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04 Oct 2019, 18:11
04 Oct 2019, 18:11
1
HarveyKlaus wrote:
The vertices of square S have coordinates (-1,-2), (-1,1), (2,1), and (2,-2) respectively. What are the coordinates of the point where the diagonals of S intersect?
a) (1/2, 1/2)
b) (1/2, -1/2)
c) (3/2, 1/2)
d) (3/2, -1/2)
e) (underoot3/2, 1/2)
a) (1/2, 1/2)
b) (1/2, -1/2)
c) (3/2, 1/2)
d) (3/2, -1/2)
e) (underoot3/2, 1/2)
Wrong OA
