Question

What are the vertices of A'B'C if ABC is dilated by a scale factor of 2?

Answer 1

The vertices of A'B'C' if A(0, 10), B(8, 6), C(4, 2) is dilated by a scale factor of 2 are;

• C. A'(0, 20), B'(16, 12), C'(8, 4)

Which method can be used to find the vertices of A'B'C'?

Given that triangle ABC is dilated by a scale factor of 2, we have;

A'B' = 2 × AB

A'C' = 2 × AC

B'C' = 2 × BC

Length of AB = √((8-0)^2 + (6-10)^2) = 4•√5

Length of AC = √((4-0)^2 + (2-10)^2) = 4•√5

Length of BC = √((8-4)^2 + (6-2)^2) = 4•√2

By multiplying the given coordinates by the scale factor, we have;

A' = 2 × (0, 10) = (0, 20)

B' = 2 × (8, 6) = (16, 12)

C' = 2 × (4, 2) = (8, 4)

A'B' = √((16-0)^2 + (12-20)^2) = 8•√5

A'C' = √((8-0)^2 + (4-20)^2) = 8•√5

B'C' = √((16-8)^2 + (12-4)^2) = 8•√2

Therefore when we have;

A'(0, 20), B'(16, 12), C'(8, 4), we get;

• A'B' = 2 × AB

1. A'C' = 2 × AC
2. B'C' = 2 × BC

The correct option is therefore;

• C. A'(0, 20), B'(16, 12), C'(8, 4)

Learn more about finding the distance between points on the coordinate plane here:

brainly.com/question/7243416

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