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)

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!
_________________

Sandy

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