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11 Dec 2017, 23:11
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B
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Question Stats:
81% (01:45) correct
19%
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The vertices of a square S have coordinates (–1, –2), (–1, 1), (2, 1) and (2, –2) respectively. What are the coordinates of the points 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) (√3/2, 1/2)
(A) (1/2, 1/2)
(B) (1/2, –1/2)
(C) (3/2, 1/2)
(D) (3/2, –1/2)
(E) (√3/2, 1/2)
12 Dec 2017, 14:47
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Bunuel
Attachment:
ddd.png [ 4.23 KiB | Viewed 3543 times ]
It does not matter which one you pick; just make sure to pick opposite, not adjacent, vertices.
Use the midpoint formula for two opposite* vertices (from diagram, D and B):
\((\frac{x_1 + x_2}{2}), (\frac{y_1 + y_2}{2})\)
\((\frac{-1 + 2}{2}), (\frac{-2 + 1}{2})\)
Intersection point: \((\frac{1}{2}\), \(-\frac{1}{2})\)
Answer B
*If you pick one vertex in Quadrant I, (x, y) is (+, +)
Go across the diagonal axis of symmetry to choose the vertex in Quadrant III, where (x,y) is (-, -)
If you pick a vertex in Quadrant II (-x, +y), choose the vertex in Quadrant IV (+x, -y)
Or sketch, which I think is faster
13 Jan 2026, 21:12
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