Answer:
The figure is not a square, because:
The diagonals DO NOT intersect at their midpoints.
The diagonals are NOT of the same length.
The diagonals are NOT perpendicular.
Step-by-step explanation:
✍️If two diagonals intersect at their midpoints, the coordinates of their midpoints will be the same.
Find the midpoints of diagonal AC and BD using the midpoint formula, [tex] M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) [/tex].
Midpoint (M) of AC, for A(-4, -6) and C(6, -18):
[tex] M(\frac{-4 + 6}{2}, \frac{-6 + (-18)}{2}) [/tex]
[tex] M(\frac{2}{2}, \frac{24}{2}) [/tex]
[tex] M(1, 12) [/tex]
Midpoint of diagonal AC = (1, 12)
Midpoint (M) of BD, for B(-12, -12) and D(13, -1):
[tex] M(\frac{-12 + 13}{2}, \frac{-12 +(-1)}{2}) [/tex]
[tex] M(\frac{1}{2}, \frac{-13}{2}) [/tex]
Midpoint of diagonal BD = [tex] M(\frac{1}{2}, \frac{-13}{2}) [/tex]
The coordinates of the midpoint of diagonal AC and diagonal BD are not the same, therefore, the diagonals do not intersect at their midpoints.
✍️Use distance formula to calculate the length of each diagonal to determine whether they are of the same length.
Distance between A(-4, -6) and C(6, -18):
[tex] AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
[tex] AC = \sqrt{(6 -(-4))^2 + (-18 -(-6))^2} [/tex]
[tex] AC = \sqrt{(10)^2 + (-12)^2} [/tex]
[tex] AC = \sqrt{100 + 144} = \sqrt{244} [/tex]
[tex] AC = 15.6 [/tex] (nearest tenth)
Distance between B(-12, -12) and D(13, -1):
[tex] BD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
[tex] BD = \sqrt{(13 - (-12))^2 + (-1 -(-12))^2} [/tex]
[tex] BD = \sqrt{(25)^2 + (11)^2} [/tex]
[tex] BD = \sqrt{625 + 121} = \sqrt{746} [/tex]
[tex] BD = 27.3 [/tex] (nearest tenth)
Diagonal AC and BD are not of the same length.
✍️If the diagonals are perpendicular, the product of their slope would equal -1.
Slope of diagonal AC:
A(-4, -6) and C(6, -18)
[tex] slope = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-18 -(-6)}{6 -(-4)} = \frac{-12}{10} = -\frac{6}{5} [/tex]
Slope of diagonal BD:
B(-12, -12) and D(13, -1)
[tex] slope = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 -(-12)}{13 - (-12)} = \frac{11}{25} [/tex]
Product of their slope:
[tex] -\frac{6}{5}*\frac{11}{25} = \frac{66}{125} [/tex]
The product of their slope doesn't equal -1. Therefore, diagonal AC and BD are not perpendicular to each other.
