You can use the fact that diagonal of a square are equal and perpendicular to each other.
The option D: Segment BD has a slope of 3 and a length of [tex]2\sqrt{10}[/tex]
One theorem in mathematics states that:
"If diagonals of a quadrilateral are equal and perpendicular to each other, then that quadrilateral is a square."
For ABCD to be square, we need AC and BD to be of equal length and perpendicular to each other.
Two lines are perpendicular if their slopes are negative reciprocal of each other.
[tex]Slope(AC) = \dfrac{rise}{run} = \dfrac{C_y - A_y}{C_x - A_x} = \dfrac{4-6}{8-2} = -\dfrac{1}{3}[/tex]
Thus, slope of BD needed to be negative reciprocal of -1/3 which is 3 for ABCD to be square (first needed condition)
[tex]|AC| = \sqrt{(C_y - A_y)^2 + (C_x - A_x)^2} = \sqrt{(4-8)^2 + (8-6)^2} = \sqrt{2^2 + 6^2} = \sqrt{40} = 2\sqrt{10}[/tex]
The second condition needs |BD| = |AC|.
Thus, length of BC should be [tex]2\sqrt{10}[/tex]
Thus, the sufficient condition for ABCD to be a square is:
Option D: Segment BD has a slope of 3 and a length of [tex]2\sqrt{10}[/tex]
Learn more about square and its diagonals here:
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