Question

3-2 lesson quiz
translations
for items 1 and 2, use △abc.

1. what are the vertices of △abc produced by ( t_{langle -3, 6

angle}(\triangle abc) = \triangle abc )?
ⓐ ( a(0, 6), b(0, 4), c(-3, 3) )
ⓑ ( a(6, 6), b(6, 4), c(3, 3) )
ⓒ ( a(0, -6), b(0, -8), c(-3, 9) )
ⓓ ( a(6, -6), b(6, -8), c(3, 9) )

1. suppose △def is the image of a translation of △abc. if d is at (-6, -2), what translation rule maps △abc to △def?

ⓐ ( t_{langle 9, 2
angle}(\triangle abc) = \triangle def )
ⓑ ( t_{langle 9, -2
angle}(\triangle abc) = \triangle def )
ⓒ ( t_{langle -9, 2
angle}(\triangle abc) = \triangle def )
ⓓ ( t_{langle -9, -2
angle}(\triangle abc) = \triangle def )

1. suppose the equation of line p is ( x = 2 ) and the equation of line q is ( x = -1 ). what translation is equivalent to ( r_p circ r_q )?

( r_p circ r_q = t_{langle m, n
angle} ) where ( m = square ) and ( n = square ).

1. what is the composition of translations ( (t_{langle -3, 4

angle} circ t_{langle 8, -7
angle})(x, y) ) written as one translation?
( (t_{langle -3, 4
angle} circ t_{langle 8, -7
angle})(x, y) = t_{langle m, n
angle}(x, y) ) where ( m = square ) and ( n = square ).

1. how many units apart are parallel lines m and n such that ( t_{langle 0, -12

angle}(x, y) = (r_n circ r_m)(x, y) )?

Explanation:

Item 1

Step1: Find original vertices

From the graph, \( A(3, 0) \), \( B(3, -2) \), \( C(0, -3) \).

Step2: Apply translation \( T_{\langle -3, 6

angle} \)
For a point \( (x, y) \), translation \( T_{\langle h, k
angle} \) gives \( (x + h, y + k) \).

• \( A' = (3 - 3, 0 + 6) = (0, 6) \)
• \( B' = (3 - 3, -2 + 6) = (0, 4) \)
• \( C' = (0 - 3, -3 + 6) = (-3, 3) \)

Step1: Find \( A \)’s coordinate

From graph, \( A(3, 0) \). \( D(-6, -2) \).

Step2: Calculate translation vector

Let translation be \( T_{\langle h, k
angle} \), so \( 3 + h = -6 \), \( 0 + k = -2 \).
\( h = -6 - 3 = -9 \), \( k = -2 - 0 = -2 \). So translation is \( T_{\langle -9, -2
angle} \).

Step1: Recall reflection over vertical lines

Reflection over \( x = a \): \( (x, y) \to (2a - x, y) \).
First, reflect over \( q: x = -1 \): \( (x, y) \to (2(-1) - x, y) = (-2 - x, y) \).
Then reflect over \( p: x = 2 \): \( (-2 - x, y) \to (2(2) - (-2 - x), y) = (4 + 2 + x, y) = (x + 6, y) \).

Step2: Identify translation

Translation \( T_{\langle 6, 0
angle} \), so \( m = 6 \), \( n = 0 \).

Answer:

A. \( A'(0, 6) \), \( B'(0, 4) \), \( C'(-3, 3) \)

Item 2