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yaroslaw [1]
2 years ago
8

A triangle has vertices at B(−3, 0), C(2, −1), D(−1, 2). Which series of transformations would produce an image with vertices B″

(4, 1), C″(−1, 0), D″(2, 3)?

Mathematics
1 answer:
vichka [17]2 years ago
7 0

We must apply the <em>transformation</em> rules (x, y) → (- x, y) and (x, y) → (x + 1, y + 1) to generate the vertices B''(x, y) = (4, 1), C''(x, y) = (- 1, 0), D''(x, y) = (2, 3). (Right choice: B)

<h3>How to determine the right transformation rules for a set of three points</h3>

In this question we have three points on a <em>Cartesian</em> plane: B(x, y) = (- 3, 0), C(x, y) = (2, - 1), D(x, y) = (- 1, 2). First, we proceed to apply the <em>transformation</em> rule (x, y) → (- x, y):

B'(x, y) = (3, 0), C'(x, y) = (- 2, -1), D'(x, y) = (1, 2)

Then, we use the <em>transformation</em> rule (x, y) → (x + 1, y + 1):

B''(x, y) = (4, 1), C''(x, y) = (- 1, 0), D''(x, y) = (2, 3)

Then, we must apply the <em>transformation</em> rules (x, y) → (- x, y) and (x, y) → (x + 1, y + 1) to generate the vertices B''(x, y) = (4, 1), C''(x, y) = (- 1, 0), D''(x, y) = (2, 3).

To learn more on transformation rules: brainly.com/question/9201867

#SPJ1

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