Answer:
D. D₂ (∆XYZ)
Explanation:
Dₙ = (x,y) → ((n)x, (n)y)
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X = (-1, 2).
Y = (2, -1).
Z = (-2, -1).
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Dₙ ∆XYZ:
X(-1, 2) → X'(-2, 4)
Y(2, -1) → Y'(4, -2)
Z(-2, -1) → Z'(-4, -2)
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-2 / -1 = 2, 4 / 2 = 2.
4 / 2 = 2, -2 / -1 = 2.
-4 / -2 = 2, -2 / -1 = 2.
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D₂ = (x,y) → ((2)x, (2)y).
X(-1, 2) → ((2)(-1), (2)(2)) = X'(-2, 4)
Y(2, -1) → ((2)(2), (2)(-1)) = Y'(4, -2)
Z(-2, -1) → ((2)(-2), (2)(-1)) = Z'(-4, -2)
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