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3 years ago

9

What transformation were applied to ABC to obtain A’B’C?

2 answers:

Cloud [144]3 years ago

6 0

Answer:

A = (3,-2)

reflect along the y -axis ; set x=-x

(-3,-2)

reflect along the x-axis ; set y=-y

A' = (-3,2)

Same for B,C and D

ABCD is reflected across the y-axis and then the x-axis.

Step-by-step explanation:

emmasim [6.3K]3 years ago

6 0

Answer:

A )rotate 90 degrees counterclockwise,then shift 2 units up.

Step-by-step explanation:

Given : graph .

To find : what transformation were applied to ABC to obtain A’B’C.

Solution : We have given graph

Parent ABC triangle with coordinates A(2 ,2) ; B (5 ,6) ; C ( 8 , 2).

Transformed triangle with coordinates A' (-2 ,4) ; B' (-6 ,7) ; C' ( -2 ,15).

By the rotational of 90° counter clockwise : (x ,y) →→ ( -y , x ).

And it is shifted by 2 unit up

(x ,y) →→ ( -y , x + 2 ).

A(2 ,2) →→ A' (-2 ,4)

B (5 ,6) →→ B' (-6 ,7)

C ( 8 , 2) →→ C' ( -2 ,15).

Therefore, A )rotate 90 degrees counterclockwise,then shift 2 units up.

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Read 2 more answers

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Step-by-step explanation:

Given;

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$M = \frac{X_1 + X_2}{2} , \frac{Y_1 + Y_2}{2}\\\\M= (1,1)\\\\1,1 =\frac{X_1 + X_2}{2} , \frac{Y_1 + Y_2}{2}\\\\ \frac{X_1 + X_2}{2} = 1 ----equation (1)\\\\\frac{Y_1 + Y_2}{2} = 1 -----equation(2)\\\\Solving \ equation(1)\\\\\frac{-4+ X_2}{2} = 1\\\\-4+ X_2 = 2\\\\X_2 = 2+4\\\\X_2 = 6\\\\Solving \ equation(2)\\\\\frac{Y_1 + Y_2}{2} = 1\\\\\frac{2+ Y_2}{2} = 1\\\\2+Y_2 = 2\\\\Y_2 = 2-2\\\\Y_2 = 0\\\\B = (X_2, \ Y_2)\\\\B = (6, \ 0)$

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