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Given a triangle ABC at points A = ( - 2, 2 ) B = ( 2, 5 ) C = ( 2, 0 ), and a first transformation of right 4 and up 3, and a second transformation of left 2 and down 5, what would be the location of the final point B'' ?

Given a triangle ABC at points A 2 2 B 2 5 C 2 0 and a first transformation of right 4 and up 3 and a second transformation of left 2 and down 5 what would be t class=

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Answer

a. (4, 3)

Step-by-step explanation

The translation of a point (x, y) a units to the right and b units up transforms the point into (x + a, y + b).

Considering point B(2, 5), translating it 4 units to the right and 3 units up, we get:

B(2, 5) → (2+4, 5+3) → B'(6, 8)

The translation of a point (x, y) c units to the left and d units down transforms the point into (x - c, y - d).

Considering point B'(6, 8), translating it 2 units to the left and 5 units down, we get:

B'(6, 8) → (6 - 2, 8 - 5) → B''(4, 3)

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