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Given the vertices, determine the most specific classification: Quadrilateral, Parallelogram, Rectangle, Rhombus or Square. Item 17 Item 18 T(-1, 2), E(7, 6), A(10, 0), M(2,-4) Item 19 (1 point) Item 20 The most specific classification for the above vertices is a

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To classify the quadrilateral formed by the points T(-1, 2), E(7, 6), A(10, 0), and M(2, -4), we need to calculate the lengths of the sides and the slopes of the sides to determine if it's a parallelogram, rectangle, rhombus, square, or just a general quadrilateral.

  1. Calculate the lengths of the sides:

    • \( TE = \sqrt{(7 - (-1))^2 + (6 - 2)^2} = \sqrt{(8)^2 + (4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \)
    • \( EA = \sqrt{(10 - 7)^2 + (0 - 6)^2} = \sqrt{(3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \)
    • \( AM = \sqrt{(2 - 10)^2 + (-4 - 0)^2} = \sqrt{(-8)^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \)
    • \( MT = \sqrt{(-1 - 2)^2 + (2 - (-4))^2} = \sqrt{(-3)^2 + (6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \)

  2. Calculate the slopes of the sides to check for parallelism:

    • Slope of \( TE = \frac{6 - 2}{7 - (-1)} = \frac{4}{8} = \frac{1}{2} \)
    • Slope of \( EA = \frac{0 - 6}{10 - 7} = \frac{-6}{3} = -2 \)
    • Slope of \( AM = \frac{-4 - 0}{2 - 10} = \frac{-4}{-8} = \frac{1}{2} \)
    • Slope of \( MT = \frac{2 - (-4)}{-1 - 2} = \frac{6}{-3} = -2 \)

Now, observe the following:

  • Sides \( TE \) and \( AM \) have equal slopes, indicating they are parallel.
  • Sides \( EA \) and \( MT \) have equal slopes, indicating they are also parallel.
  • We have parallel opposite sides, confirming that it is a parallelogram.

Next, we check:

  • The lengths of the opposite sides are equal: \( TE = AM \) and \( EA = MT \).
  • Notably, \( TE \) and \( AM \) are equal, and \( EA \) and \( MT \) are also equal.
  1. Check for rectangles and squares:
    • A rectangle would require at least one pair of adjacent sides to be perpendicular.
    • Check if adjacent sides \( TE \) and \( EA \) are perpendicular:
      • The product of their slopes should equal -1:
      • Slope of \( TE \) is \( \frac{1}{2} \) and the slope of \( EA \) is \( -2 \).
      • \( \frac{1}{2} \times (-2) = -1 \), hence they are perpendicular.

Since we have two pairs of equal-length sides and one pair of adjacent sides that are perpendicular:

  • This confirms that it is also a rectangle.
  1. Check for rhombus and square:
    • A rhombus requires all sides to be equal, which they are not in this case.
    • A square requires all sides to be equal and for it to be a rectangle. Since not all sides equal, it is not a square.

Thus, the most specific classification is a rectangle.