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Given the vertices, determine the most specific classification: Quadrilateral, Parallelogram, Rectangle, Rhombus or Square. Q(−...Question
Given the vertices, determine the most specific classification: Quadrilateral, Parallelogram, Rectangle, Rhombus or Square. Q(−8,1), U(3,5), A(7,−1), Y(−4,−4) (1 point) The most specific classification for the above vertices is a
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Answered by
GPT-4o mini
To classify the quadrilateral formed by the points Q(−8,1), U(3,5), A(7,−1), and Y(−4,−4), we need to check the properties of the shape based on the coordinates of the vertices.
1. **Calculate the lengths of the sides**:
- \( \text{Distance between Q and U} \):
\[
\sqrt{(3 - (-8))^2 + (5 - 1)^2} = \sqrt{(3 + 8)^2 + (5 - 1)^2} = \sqrt{11^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137}
\]
- \( \text{Distance between U and A} \):
\[
\sqrt{(7 - 3)^2 + (-1 - 5)^2} = \sqrt{(4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52}
\]
- \( \text{Distance between A and Y} \):
\[
\sqrt{(-4 - 7)^2 + (-4 - (-1))^2} = \sqrt{(-11)^2 + (-4 + 1)^2} = \sqrt{121 + 9} = \sqrt{130}
\]
- \( \text{Distance between Y and Q} \):
\[
\sqrt{(-8 - (-4))^2 + (1 - (-4))^2} = \sqrt{(-8 + 4)^2 + (1 + 4)^2} = \sqrt{(-4)^2 + (5)^2} = \sqrt{16 + 25} = \sqrt{41}
\]
2. **Check for parallel sides** (to determine if it's a parallelogram):
- Calculate the slopes of the sides:
- Slope of QU:
\[
m_{QU} = \frac{5 - 1}{3 - (-8)} = \frac{4}{11}
\]
- Slope of UA:
\[
m_{UA} = \frac{-1 - 5}{7 - 3} = \frac{-6}{4} = -\frac{3}{2}
\]
- Slope of AY:
\[
m_{AY} = \frac{-4 - (-1)}{-4 - 7} = \frac{-3}{-11} = \frac{3}{11}
\]
- Slope of YQ:
\[
m_{YQ} = \frac{1 - (-4)}{-8 - (-4)} = \frac{5}{-4} = -\frac{5}{4}
\]
We see that the slopes do not match in pairs. Therefore, there are no pairs of parallel sides.
3. **Conclusion**:
Since the quadrilateral does not have any pairs of parallel sides, it cannot qualify as a parallelogram, rectangle, rhombus, or square.
Thus, the most specific classification for the shape formed by the vertices Q(−8,1), U(3,5), A(7,−1), and Y(−4,−4) is a **Quadrilateral**.
1. **Calculate the lengths of the sides**:
- \( \text{Distance between Q and U} \):
\[
\sqrt{(3 - (-8))^2 + (5 - 1)^2} = \sqrt{(3 + 8)^2 + (5 - 1)^2} = \sqrt{11^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137}
\]
- \( \text{Distance between U and A} \):
\[
\sqrt{(7 - 3)^2 + (-1 - 5)^2} = \sqrt{(4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52}
\]
- \( \text{Distance between A and Y} \):
\[
\sqrt{(-4 - 7)^2 + (-4 - (-1))^2} = \sqrt{(-11)^2 + (-4 + 1)^2} = \sqrt{121 + 9} = \sqrt{130}
\]
- \( \text{Distance between Y and Q} \):
\[
\sqrt{(-8 - (-4))^2 + (1 - (-4))^2} = \sqrt{(-8 + 4)^2 + (1 + 4)^2} = \sqrt{(-4)^2 + (5)^2} = \sqrt{16 + 25} = \sqrt{41}
\]
2. **Check for parallel sides** (to determine if it's a parallelogram):
- Calculate the slopes of the sides:
- Slope of QU:
\[
m_{QU} = \frac{5 - 1}{3 - (-8)} = \frac{4}{11}
\]
- Slope of UA:
\[
m_{UA} = \frac{-1 - 5}{7 - 3} = \frac{-6}{4} = -\frac{3}{2}
\]
- Slope of AY:
\[
m_{AY} = \frac{-4 - (-1)}{-4 - 7} = \frac{-3}{-11} = \frac{3}{11}
\]
- Slope of YQ:
\[
m_{YQ} = \frac{1 - (-4)}{-8 - (-4)} = \frac{5}{-4} = -\frac{5}{4}
\]
We see that the slopes do not match in pairs. Therefore, there are no pairs of parallel sides.
3. **Conclusion**:
Since the quadrilateral does not have any pairs of parallel sides, it cannot qualify as a parallelogram, rectangle, rhombus, or square.
Thus, the most specific classification for the shape formed by the vertices Q(−8,1), U(3,5), A(7,−1), and Y(−4,−4) is a **Quadrilateral**.
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