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A quadrilateral PQRS with vertices P(-2, 1), Q(0, 3), R(2, 2), and S(1, 0) is transformed by reflection in the x-axis, followed by a translation, and then a rotation of 180 degrees about the origin. F

Mathematics

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Step-by-step solution

5 steps

Here are the steps for the composite transformation:

  1. Reflect quadrilateral PQRSPQRS in the x-axis to obtain P′Q′R′S′P'Q'R'S'.

    The coordinate rule for reflection in the x-axis is (x,y)→(x,−y)(x, y) \to (x, -y).
    Given vertices P(−2,1)P(-2, 1), Q(0,3)Q(0, 3), R(2,2)R(2, 2), and S(1,0)S(1, 0):

    P′=(−2,−1)P' = (-2, -1) Q′=(0,−3)Q' = (0, -3) R′=(2,−2)R' = (2, -2) S′=(1,0)S' = (1, 0)

    The coordinates of P′Q′R′S′P'Q'R'S' are P′(−2,−1)P'(-2, -1), Q′(0,−3)Q'(0, -3), R′(2,−2)R'(2, -2), and S′(1,0)S'(1, 0).

  2. Translate P′Q′R′S′P'Q'R'S' using the vector to obtain P′′Q′′R′′S′′P''Q''R''S''.

    The translation vector is missing from the problem statement. For demonstration purposes, I will assume a translation vector of (11)\begin{pmatrix} 1 \\ 1 \end{pmatrix}.
    The coordinate rule for this translation is (x,y)→(x+1,y+1)(x, y) \to (x+1, y+1).
    Applying the rule to the vertices P′(−2,−1)P'(-2, -1), Q′(0,−3)Q'(0, -3), R′(2,−2)R'(2, -2), and S′(1,0)S'(1, 0):

    P′′=(−2+1,−1+1)=(−1,0)P'' = (-2+1, -1+1) = (-1, 0) Q′′=(0+1,−3+1)=(1,−2)Q'' = (0+1, -3+1) = (1, -2) R′′=(2+1,−2+1)=(3,−1)R'' = (2+1, -2+1) = (3, -1) S′′=(1+1,0+1)=(2,1)S'' = (1+1, 0+1) = (2, 1)

    The coordinates of P′′Q′′R′′S′′P''Q''R''S'' are P′′(−1,0)P''(-1, 0), Q′′(1,−2)Q''(1, -2), R′′(3,−1)R''(3, -1), and S′′(2,1)S''(2, 1).

  3. Rotate P′′Q′′R′′S′′P''Q''R''S'' through 180∘180^\circ about the origin to obtain P′′′Q′′′R′′′S′′′P'''Q'''R'''S'''.

    The coordinate rule for a 180∘180^\circ rotation about the origin is (x,y)→(−x,−y)(x, y) \to (-x, -y).
    Applying the rule to the vertices P′′(−1,0)P''(-1, 0), Q′′(1,−2)Q''(1, -2), R′′(3,−1)R''(3, -1), and S′′(2,1)S''(2, 1):

    P′′′=(−(−1),−(0))=(1,0)P''' = (-(-1), -(0)) = (1, 0) Q′′′=(−(1),−(−2))=(−1,2)Q''' = (-(1), -(-2)) = (-1, 2) R′′′=(−(3),−(−1))=(−3,1)R''' = (-(3), -(-1)) = (-3, 1) S′′′=(−(2),−(1))=(−2,−1)S''' = (-(2), -(1)) = (-2, -1)

    The coordinates of P′′′Q′′′R′′′S′′′P'''Q'''R'''S''' are P′′′(1,0)P'''(1, 0), Q′′′(−1,2)Q'''(-1, 2), R′′′(−3,1)R'''(-3, 1), and S′′′(−2,−1)S'''(-2, -1).

  4. Enlarge P′′′Q′′′R′′′S′′′P'''Q'''R'''S''' with a scale factor of k=12k = \frac{1}{2} centred at the origin to obtain P′′′′Q′′′′R′′′′S′′′′P''''Q''''R''''S''''.

    The coordinate rule for enlargement with scale factor k=12k=\frac{1}{2} centred at the origin is (x,y)→(12x,12y)(x, y) \to (\frac{1}{2}x, \frac{1}{2}y).
    Applying the rule to the vertices P′′′(1,0)P'''(1, 0), Q′′′(−1,2)Q'''(-1, 2), R′′′(−3,1)R'''(-3, 1), and S′′′(−2,−1)S'''(-2, -1):

    P′′′′=(12×1,12×0)=(12,0)P'''' = (\frac{1}{2} \times 1, \frac{1}{2} \times 0) = (\frac{1}{2}, 0) Q′′′′=(12×−1,12×2)=(−12,1)Q'''' = (\frac{1}{2} \times -1, \frac{1}{2} \times 2) = (-\frac{1}{2}, 1) R′′′′=(12×−3,12×1)=(−32,12)R'''' = (\frac{1}{2} \times -3, \frac{1}{2} \times 1) = (-\frac{3}{2}, \frac{1}{2}) S′′′′=(12×−2,12×−1)=(−1,−12)S'''' = (\frac{1}{2} \times -2, \frac{1}{2} \times -1) = (-1, -\frac{1}{2})

    The coordinates of P′′′′Q′′′′R′′′′S′′′′P''''Q''''R''''S'''' are P′′′′(12,0)P''''(\frac{1}{2}, 0), Q′′′′(−12,1)Q''''(-\frac{1}{2}, 1), R′′′′(−32,12)R''''(-\frac{3}{2}, \frac{1}{2}), and S′′′′(−1,−12)S''''(-1, -\frac{1}{2}).

    To create the labeled graph, plot the original quadrilateral PQRSPQRS and all the transformed images P′Q′R′S′P'Q'R'S', P′′Q′′R′′S′′P''Q''R''S'', P′′′Q′′′R′′′S′′′P'''Q'''R'''S''', and P′′′′Q′′′′R′′′′S′′′′P''''Q''''R''''S'''' using the coordinates calculated above on a coordinate plane.

  5. Compare the area relationship between PQRSPQRS and P′′′′Q′′′′R′′′′S′′′′P''''Q''''R''''S''''.

    Reflection, translation, and rotation are rigid transformations, meaning they preserve the area of the shape (the area scale factor is 1).
    Enlargement with a scale factor kk changes the area by a factor of k2k^2.
    In this sequence of transformations, the only transformation that changes the area is the final enlargement with a scale factor of k=12k = \frac{1}{2}.
    Therefore, the area of the final image P′′′′Q′′′′R′′′′S′′′′P''''Q''''R''''S'''' is k2k^2 times the area of the original quadrilateral PQRSPQRS.

    Area(P′′′′Q′′′′R′′′′S′′′′)=(12)2×Area(PQRS)\text{Area}(P''''Q''''R''''S'''') = \left(\frac{1}{2}\right)^2 \times \text{Area}(PQRS) Area(P′′′′Q′′′′R′′′′S′′′′)=14×Area(PQRS)\text{Area}(P''''Q''''R''''S'''') = \frac{1}{4} \times \text{Area}(PQRS)

    The area of P′′′′Q′′′′R′′′′S′′′′P''''Q''''R''''S'''' is one-fourth the area of PQRSPQRS.

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