Given ∆ABC with vertices at A(-2, 2), B(2, 5), and C(2, 0) and a first transformation of 4 units to the right and 4 units up, and a second transformation of 2 units to the left and 5 units down. What would be the location of the final B''?
Question
Given ∆ABC with vertices at A(-2, 2), B(2, 5), and C(2, 0) and a first transformation of 4 units to the right and 4 units up, and a second transformation of 2 units to the left and 5 units down. What would be the location of the final B''?
Solution
The first transformation moves point B(2, 5) 4 units to the right and 4 units up. This results in a new point B'(6, 9).
The second transformation moves point B'(6, 9) 2 units to the left and 5 units down. This results in the final point B''(4, 4).
Similar Questions
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