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The vertices of a square S have coordinates (–1, –2), (–1, 1), (2, 1)

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Joined: 02 Sep 2009
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The vertices of a square S have coordinates (–1, –2), (–1, 1), (2, 1)  [#permalink]

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11 Dec 2017, 22:11
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Difficulty:

25% (medium)

Question Stats:

78% (01:51) correct 22% (01:46) wrong based on 37 sessions

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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)

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The vertices of a square S have coordinates (–1, –2), (–1, 1), (2, 1)  [#permalink]

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12 Dec 2017, 13:47
Bunuel wrote:
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)

Attachment:

ddd.png [ 4.23 KiB | Viewed 564 times ]

The diagonals of a square bisect each other, so the coordinates of their intersection are the midpoint of either diagonal.

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: $$(\frac{1}{2}$$, $$-\frac{1}{2})$$

*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
The vertices of a square S have coordinates (–1, –2), (–1, 1), (2, 1) &nbs [#permalink] 12 Dec 2017, 13:47
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