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)
<h3>Which method can be used to find the vertices of A'B'C'?</h3>
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'C' = 2 × AC
- 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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(Interchange the x and y-values)
(Solve for y)
