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Question

The vertices of square ABCD are A(−5, 0), B(−1, −2), C(−3, −6) and D(−7, −4). Which of the following shows that its diagonals are congruent perpendicular bisectors of each other?

Answer 1

Show that segments AC and BD have the same midpoint.

M_ac= [(-5 - 3)/2 , (0 - 6)/2] = (-4, -3)

M_bd = [(-1 - 7)/2, (-2 - 4)/2] = (-4, -3)

Since the diagonals have the same midpoint, they bisect each other.

Show that the segments have slopes that are negative reciprocals:

m_ac = [(0 - (-6))/( -5 - (-3)) = 6/-2 = -3

m_bd = [(-2 - (-4))/(-1 - (-7))] = 2/6 = 1/3

(-3)(1/3) = -1

Since the slopes are negative reciprocals, the lines containing the diagonals are perpendicular.

For additional confirmation you could also show that the distances from each vertex to the midpoint are equal.

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