Triangle ABC has vertices at A(-3, 1), B(2, -4), and C(1, 5). It is translated 2 units right and 3 units down to form triangle A’B’C’. A'B'C' is then reflected over the line y = x to form triangle A''B''C''. What are the vertices of triangle A’'B’'C’'? A. A''(2, 1), B''(7,-4), C''(-2, 3) B. A''(2, -1), B''(7,4), C''(-2, -3) C. A''(-1, -2), B''(4,-7), C''(-3, 2) D. A''(-2, -1), B''(-7,4), C''(2, 3)

D. A''(-2, -1), B''(-7,4), C''(2, 3)

Step-by-step explanation:

Given a triangle ABC with coordinate of its vertices at A(-3, 1), B(2, -4), and C(1, 5), if it is translated 2 units right and 3 units down, the new coordinates A', B' and C' will be gotten by adding the coordinate (2, -3) to the coordinates of the original triangle as shown:

A' = (2,-3)+A(-3,1)

A' = {(2-3), (-3+1)}

A' = (-1, -2)

B' = (2,-3)+B(2, -4)

B' = {(2+2), (-3-4)}

B' = (4, -7)

C' = (2,-3)+C(1,5)

C'= {(2+1), (-3+5)}

C' = (3, 2)

If A'B'C' is now reflected over the line y = x, this means that the x coordinates will be switched with y coordinates of the triangle A'B'C'.

If A' = (-1,-2), A'' = (-2,-1)

If B' = (4,-7), B'' = (-7, 4)

If C' = (3, 2), C'' = (2, 3)

Option D is correct

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i think its 7.6

if so your welcome