Explanation Given: The Δ A ″ B ″ C ″ with A ″ ( 0 , 0 ) , B ″ ( 1 , 5 ) and C ″ ( 2 − , 7 ) is the image of Δ A B C under the translation ( x , y ) → ( x + 2 , y − 1 ) followed by the translation ( x , y ) → ( x + 1 , y + 3 ) . Approach: Translation: It is a function from the plane to the plane such that the image of the point with coordinates ( x , y ) is ( x + a , y + b ) , where a and b are real numbers. Calculation: Consider the Δ A ″ B ″ C ″ with A ″ ( 0 , 0 ) , B ″ ( 1 , 5 ) and C ″ ( 2 − , 7 ) . Also, Δ A ″ B ″ C ″ is the image of Δ A B C . The translations are given by ( x , y ) → ( x + 2 , y − 1 ) … ( 1 ) Followed by the translation ( x , y ) → ( x + 1 , y + 3 ) … ( 2 ) The coordinates of A ″ is ( 0 , 0 ) . The coordinates of point A for the first translation given by equation ( 1 ) can be found as follows: x ′ = x + 2 and y ′ = y − 1 For the second translation given by ( 2 ) is: x ″ = x ′ + 1 and y ″ = y ′ + 3 Substitute the values of x ′ and y ′ in the above equation. x ″ = ( x + 2 ) + 1 and y ″ = ( y − 1 ) + 3 x ″ = x + 2 + 1 and y ″ = y − 1 + 3 x ″ = x + 3 and y ″ = y + 2 x = x ″ − 3 and y = y ″ − 2 So, the coordinates of A is ( 3 − , 2 − ) . The coordinates of B ″ is ( 1 , 5 ) . The coordinates of point B for the first translation given by equation ( 1 ) can be found as follows: x ′ + 1 = x + 2 and y ′ + 5 = y − 1 x ′ = x + 2 − 1 and y ′ = y − 1 − 5 x ′ = x + 1 and y ′ = y − 6 For the second translation given by ( 2 ) is: x ″ = x ′ + 1 and y ″ = y ′ + 3 Substitute the values of x ′ and y ′ in the above equation To determine b. To find: The translation under which Δ A ″ B ″ C ″ is the image of Δ A B C .

12th Edition

ISBN: 9780137657988

Author: BILLSTEIN

Publisher: INTER PEAR

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Chapter 13.CR, Problem 12CR

To determine

a.

To find:

The coordinates A, B, and C of the triangle ABC under the translation (x,y)→(x+2,y−1) followed by the translation (x,y)→(x+1,y+3).

To determine

b.

To find:

The translation under which ΔA″B″C″ is the image of ΔABC.

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Chapter 13.CR, Problem 11CR

Chapter 13.CR, Problem 13CR

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Chapter 13 Solutions

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Ch. 13.1 - Prob. 13MC Ch. 13.1 - Prob. 14MC Ch. 13.1 - Prob. 15MC Ch. 13.1 - Prob. 16MC Ch. 13.1 - Prob. 19MC Ch. 13.1 - Prob. 1NAEP Ch. 13.1A - For each of the following, find the image of the...Ch. 13.1A - Construct the image BC under the translation...Ch. 13.1A - Find the coordinates of the image for each of the...Ch. 13.1A - Prob. 4A Ch. 13.1A - Prob. 7A Ch. 13.1A - Prob. 8A Ch. 13.1A - If y=2x+3 is the image of the line k under the...Ch. 13.1A - Prob. 10A Ch. 13.1A - Prob. 11A Ch. 13.1A - Prob. 12A Ch. 13.1A - a. Draw a line l and any two points A and B so...Ch. 13.1A - Prob. 14A Ch. 13.1A - Prob. 15A Ch. 13.1A - Prob. 16A Ch. 13.1A - Prob. 18A Ch. 13.1A - a. 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a. To find: The coordinates A , B , and C of the triangle ABC under the translation ( x , y ) → ( x + 2 , y − 1 ) followed by the translation ( x , y ) → ( x + 1 , y + 3 ) .

Solution Summary: The author calculates the coordinates of A, B, and C of the triangle ABC under the translation (x,y)

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a.

To find:

The coordinates A, B, and C of the triangle ABC under the translation (x,y)→(x+2,y−1) followed by the translation (x,y)→(x+1,y+3).

b.

To find:

The translation under which ΔA″B″C″ is the image of ΔABC.

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Chapter 13 Solutions

a. To find: The coordinates A , B , and C of the triangle ABC under the translation ( x , y ) → ( x + 2 , y − 1 ) followed by the translation ( x , y ) → ( x + 1 , y + 3 ) .

Solution Summary: The author calculates the coordinates of A, B, and C of the triangle ABC under the translation (x,y)

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a. To find: The coordinates A, B, and C of the triangle ABC under the translation (x,y)→(x+2,y−1) followed by the translation (x,y)→(x+1,y+3).

b. To find: The translation under which ΔA″B″C″ is the image of ΔABC.

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Chapter 13 Solutions

a.

To find:

The coordinates A, B, and C of the triangle ABC under the translation (x,y)→(x+2,y−1) followed by the translation (x,y)→(x+1,y+3).

b.

To find:

The translation under which ΔA″B″C″ is the image of ΔABC.